Some differential subordinations using Ruscheweyh derivative and Sălăgean operator

Abstract

In the present paper we study the operator defined by using the Ruscheweyh derivative $R^{n}f(z)$ and the Sălăgean operator $S^{n}f(z)$, denoted by $L_{\alpha }^n:\mathcal{A}\to \mathcal{A}$, $L_{\alpha }^{n}f(z)=(1-\alpha )R^{n}f(z)+\alpha S^{n}f(z)$, $z\in U$, 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 with ${\mathcal{A}}_1=\mathcal{A}$. We obtain several differential subordinations regarding the operator $L_{\alpha }^n$.

MSC:30C45, 30A20, 34A40.

1 Introduction

Denote by U the unit disc of the complex plane, $U=\{z\in \mathbb{C}:|z|<1\}$, and by $\mathcal{H}(U)$ the space of holomorphic functions in U. Let ${\mathcal{A}}_n=\{f\in \mathcal{H}(U):f(z)=z+a_{n+1}{z}^{n+1}+\cdots ,z\in U\}$ with ${\mathcal{A}}_1=\mathcal{A}$ 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 $K=\{f\in \mathcal{A}:Re\frac{z{f}^{\mathrm{\prime }\mathrm{\prime }}(z)}{f^{\mathrm{\prime }}(z)}+1>0,z\in U\}$ 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 a univalent function in U. If p is analytic in U and satisfies the (second-order) differential subordination

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

(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 (Sălăgean [1])

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

$$\begin{array}{r}{S}^{0}f(z)=f(z),\\ S^{1}f(z)=z{f}^{\mathrm{\prime }}(z),\\ \cdots \\ S^{n+1}f(z)=z{(S^{n}f(z))}^{\mathrm{\prime }},z\in U.\end{array}$$

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

Definition 1.2 (Ruscheweyh [2])

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

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

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

Definition 1.3 ([3])

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

$$L_{\alpha }^{n}f(z)=(1-\alpha )R^{n}f(z)+\alpha S^{n}f(z),z\in U.$$

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

This operator was studied also in [3–5].

Lemma 1.1 (Hallenbeck and Ruscheweyh [[6], 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

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

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 [6])

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

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

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

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

$${\left(\frac{L_{\alpha }^{n}f(z)}{z}\right)}^{\delta -1}{(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}\prec h(z),z\in U,$$

(2.1)

then

$${\left(\frac{L_{\alpha }^{n}f(z)}{z}\right)}^{\delta }\prec g(z),z\in U,$$

and this result is sharp.

Proof By using the properties of the operator $L_{\alpha }^n$, we have

$$L_{\alpha }^{n}f(z)=z+\sum _{j=2}^{\mathrm{\infty }}(\alpha j^n+(1-\alpha )C_{n+j-1}^n)a_j{z}^j,z\in U.$$

Consider $p(z)={(\frac{L_{\alpha }^{n}f(z)}{z})}^{\delta }={(\frac{z+{\sum }_{j=2}^{\mathrm{\infty }}(\alpha j^n+(1-\alpha )C_{n+j-1}^n)a_j{z}^j}{z})}^{\delta }=1+p_{\delta }{z}^{\delta }+p_{\delta +1}{z}^{\delta +1}+\cdots$ , $z\in U$.

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

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

Then (2.1) becomes

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

By using Lemma 1.2, we have

$$p(z)\prec g(z),z\in U,\mathit{\text{i.e.}}\text{,}{\left(\frac{L_{\alpha }^{n}f(z)}{z}\right)}^{\delta }\prec g(z),z\in U.$$

 □

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

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

$${\left(\frac{L_{\alpha }^{n}f(z)}{z}\right)}^{\delta -1}{(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}\prec h(z),z\in U,$$

(2.2)

then

$${\left(\frac{L_{\alpha }^{n}f(z)}{z}\right)}^{\delta }\prec q(z),z\in U,$$

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

Proof Let

$$\begin{array}{rcl}p(z)& =& {\left(\frac{L_{\alpha }^{n}f(z)}{z}\right)}^{\delta }={\left(\frac{z+{\sum }_{j=2}^{\mathrm{\infty }}(\alpha j^n+(1-\alpha )C_{n+j-1}^n)a_j{z}^j}{z}\right)}^{\delta }\\ =& {(1+\sum _{j=2}^{\mathrm{\infty }}(\alpha j^n+(1-\alpha )C_{n+j-1}^n)a_j{z}^{j-1})}^{\delta }=1+\sum _{j=\delta +1}^{\mathrm{\infty }}{p}_j{z}^{j-1}\end{array}$$

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

Differentiating, we obtain ${(\frac{L_{\alpha }^{n}f(z)}{z})}^{\delta -1}{(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}=p(z)+\frac{1}{\delta }z{p}^{\mathrm{\prime }}(z)$, $z\in U$, and (2.2) becomes

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

Using Lemma 1.1, we have

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

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

$${\left(\frac{L_{\alpha }^{n}f(z)}{z}\right)}^{\delta -1}{(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}\prec h(z),z\in U,$$

(2.3)

then

$${\left(\frac{L_{\alpha }^{n}f(z)}{z}\right)}^{\delta }\prec q(z),z\in U,$$

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

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

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

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

 □

Remark 2.1 For $n=1$, $\alpha =2$, $\delta =1$, we obtain the same example as in [[7], Example 2.2.1, p.26].

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

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

$$\frac{(\gamma +1)z}{\gamma }\frac{L_{\alpha }^{n}f(z)}{{(L_{\alpha }^{n+1}f(z))}^2}+\frac{z^2}{\gamma }\frac{L_{\alpha }^{n}f(z)}{{(L_{\alpha }^{n+1}f(z))}^2}[\frac{{(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}}{L_{\alpha }^{n}f(z)}-2\frac{{(L_{\alpha }^{n+1}f(z))}^{\mathrm{\prime }}}{L_{\alpha }^{n+1}f(z)}]\prec h(z),z\in U,$$

(2.4)

holds, then

$$z\frac{L_{\alpha }^{n}f(z)}{{(L_{\alpha }^{n+1}f(z))}^2}\prec g(z),z\in U,$$

and this result is sharp.

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

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

Relation (2.4) becomes

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

By using Lemma 1.2, we have

$$p(z)\prec g(z),z\in U,\mathit{\text{i.e.}}\text{,}z\frac{L_{\alpha }^{n}f(z)}{{(L_{\alpha }^{n+1}f(z))}^2}\prec g(z),z\in U.$$

 □

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

If $\alpha \ge 0$, $\gamma \in \mathbb{C}\mathrm{\setminus }\{0\}$ is a complex number with $Re\gamma \ge 0$, $n\in \mathbb{N}$, $f\in \mathcal{A}$ and satisfies the differential subordination

$$\frac{(\gamma +1)z}{\gamma }\frac{L_{\alpha }^{n}f(z)}{{(L_{\alpha }^{n+1}f(z))}^2}+\frac{z^2}{\gamma }\frac{L_{\alpha }^{n}f(z)}{{(L_{\alpha }^{n+1}f(z))}^2}[\frac{{(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}}{L_{\alpha }^{n}f(z)}-2\frac{{(L_{\alpha }^{n+1}f(z))}^{\mathrm{\prime }}}{L_{\alpha }^{n+1}f(z)}]\prec h(z),z\in U,$$

(2.5)

then

$$z\frac{L_{\alpha }^{n}f(z)}{{(L_{\alpha }^{n+1}f(z))}^2}\prec q(z),z\in U,$$

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

Proof Let $p(z)=z\frac{L_{\alpha }^{n}f(z)}{{(L_{\alpha }^{n+1}f(z))}^2}$, $z\in U$, $p\in \mathcal{H}[1,1]$. Differentiating, we obtain $p(z)+\frac{z}{\gamma }{p}^{\mathrm{\prime }}(z)=\frac{(\gamma +1)z}{\gamma }\frac{L_{\alpha }^{n}f(z)}{{(L_{\alpha }^{n+1}f(z))}^2}+\frac{z^2}{\gamma }\frac{L_{\alpha }^{n}f(z)}{{(L_{\alpha }^{n+1}f(z))}^2}[\frac{{(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}}{L_{\alpha }^{n}f(z)}-2\frac{{(L_{\alpha }^{n+1}f(z))}^{\mathrm{\prime }}}{L_{\alpha }^{n+1}f(z)}]$, $z\in U$, and (2.5) becomes

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

Using Lemma 1.1, we have

$$p(z)\prec q(z),z\in U,\mathit{\text{i.e.}}\text{,}z\frac{L_{\alpha }^{n}f(z)}{{(L_{\alpha }^{n+1}f(z))}^2}\prec q(z)=\frac{\gamma }{z^{\gamma }}{\int }_0^{z}h(t)t^{\gamma -1}dt,z\in U,$$

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 $h(z)=g(z)+\frac{z}{\gamma }{g}^{\mathrm{\prime }}(z)$, $z\in U$, where $\gamma >0$.

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

$$\frac{(\gamma +2)z^2}{\gamma }\frac{{(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}}{L_{\alpha }^{n}f(z)}+\frac{z^3}{\gamma }[\frac{{(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }\mathrm{\prime }}}{L_{\alpha }^{n}f(z)}-{\left(\frac{{(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}}{L_{\alpha }^{n}f(z)}\right)}^2]\prec h(z),z\in U,$$

(2.6)

holds, then

$$z^2\frac{{(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}}{L_{\alpha }^{n}f(z)}\prec g(z),z\in U.$$

This result is sharp.

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

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

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

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

By using Lemma 1.2, we have

$$p(z)\prec g(z),z\in U,\mathit{\text{i.e.}}\text{,}{z}^2\frac{{(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}}{L_{\alpha }^{n}f(z)}\prec g(z),z\in U,$$

and this result is sharp. □

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

If $\alpha \ge 0$, $\gamma \in \mathbb{C}\mathrm{\setminus }\{0\}$ is a complex number with $Re\gamma \ge 0$, $n\in \mathbb{N}$, $f\in \mathcal{A}$ and satisfies the differential subordination

$$\frac{(\gamma +2)z^2}{\gamma }\frac{{(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}}{L_{\alpha }^{n}f(z)}+\frac{z^3}{\gamma }[\frac{{(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }\mathrm{\prime }}}{L_{\alpha }^{n}f(z)}-{\left(\frac{{(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}}{L_{\alpha }^{n}f(z)}\right)}^2]\prec h(z),z\in U,$$

(2.7)

then

$$z^2\frac{{(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}}{L_{\alpha }^{n}f(z)}\prec q(z),z\in U,$$

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

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

Differentiating, we obtain $p(z)+\frac{z}{\gamma }{p}^{\mathrm{\prime }}(z)=\frac{(\gamma +2)z^2}{\gamma }\frac{{(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}}{L_{\alpha }^{n}f(z)}+\frac{z^3}{\gamma }[\frac{{(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }\mathrm{\prime }}}{L_{\alpha }^{n}f(z)}-{(\frac{{(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}}{L_{\alpha }^{n}f(z)})}^2]$, $z\in U$, and (2.7) becomes

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

Using Lemma 1.1, we have

$$p(z)\prec q(z),z\in U,\mathit{\text{i.e.}}\text{,}{z}^2\frac{{(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}}{L_{\alpha }^{n}f(z)}\prec q(z)=\frac{\gamma }{z^{\gamma }}{\int }_0^{z}h(t)t^{\gamma -1}dt,z\in U,$$

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

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

$$1-\frac{L_{\alpha }^{n}f(z)\cdot {(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }\mathrm{\prime }}}{{[{(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}]}^2}\prec h(z),z\in U,$$

(2.8)

holds, then

$$\frac{L_{\alpha }^{n}f(z)}{z{(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}}\prec g(z),z\in U.$$

This result is sharp.

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

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

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

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

By using Lemma 1.2, we have

$$p(z)\prec g(z),z\in U,\mathit{\text{i.e.}}\text{,}\frac{L_{\alpha }^{n}f(z)}{z{(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}}\prec g(z),z\in U,$$

and this result is sharp. □

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

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

$$1-\frac{L_{\alpha }^{n}f(z)\cdot {(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }\mathrm{\prime }}}{{[{(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}]}^2}\prec h(z),z\in U,$$

(2.9)

then

$$\frac{L_{\alpha }^{n}f(z)}{z{(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}}\prec q(z),z\in U,$$

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

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

Differentiating, we obtain $1-\frac{L_{\alpha }^{n}f(z)\cdot {(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }\mathrm{\prime }}}{{[{(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}]}^2}=p(z)+z{p}^{\mathrm{\prime }}(z)$, $z\in U$, and (2.9) becomes

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

Using Lemma 1.1, we have

$$p(z)\prec q(z),z\in U,\mathit{\text{i.e.}}\text{,}\frac{L_{\alpha }^{n}f(z)}{z{(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}}\prec q(z)=\frac{1}{z}{\int }_0^{z}h(t)dt,z\in U,$$

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

$$1-\frac{L_{\alpha }^{n}f(z)\cdot {(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }\mathrm{\prime }}}{{[{(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}]}^2}\prec h(z),z\in U,$$

(2.10)

then

$$\frac{L_{\alpha }^{n}f(z)}{z{(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}}\prec q(z),z\in U,$$

where q is given by $q(z)=(2\beta -1)+2(1-\beta )\frac{ln(1+z)}{z}$, $z\in U$. 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 $p(z)=\frac{L_{\alpha }^{n}f(z)}{z{(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}}$, the differential subordination (2.10) becomes

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

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

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

 □

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

Let $f(z)=z+z^2$, $z\in U$. For $n=1$, $\alpha =2$, we obtain $L_2^{1}f(z)=-R^{1}f(z)+2S^{1}f(z)=-z{f}^{\mathrm{\prime }}(z)+2z{f}^{\mathrm{\prime }}(z)=z{f}^{\mathrm{\prime }}(z)=z+2z^2$.

Then ${(L_2^{1}f(z))}^{\mathrm{\prime }}=1+4z$,

$$\begin{array}{c}\frac{L_2^{1}f(z)}{z{(L_2^{1}f(z))}^{\mathrm{\prime }}}=\frac{z+2z^2}{z(1+4z)}=\frac{1+2z}{1+4z},\hfill \\ 1-\frac{L_2^{1}f(z)\cdot {(L_2^{1}f(z))}^{\mathrm{\prime }\mathrm{\prime }}}{{[{(L_2^{1}f(z))}^{\mathrm{\prime }}]}^2}=1-\frac{(z+2z^2)\cdot 4}{{(1+4z)}^2}=\frac{8z^2+4z+1}{{(1+4z)}^2}.\hfill \end{array}$$

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

$$\frac{8z^2+4z+1}{{(1+4z)}^2}\prec \frac{1-z}{1+z},z\in U,$$

induce

$$\frac{1+2z}{1+4z}\prec -1+\frac{2ln(1+z)}{z},z\in U.$$

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

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

$${\left[{(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}\right]}^2+L_{\alpha }^{n}f(z)\cdot {(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }\mathrm{\prime }}\prec h(z),z\in U,$$

(2.11)

holds, then

$$\frac{L_{\alpha }^{n}f(z)\cdot {(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}}{z}\prec g(z),z\in U.$$

This result is sharp.

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

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

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

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

By using Lemma 1.2, we have

$$p(z)\prec g(z),z\in U,\mathit{\text{i.e.}}\text{,}\frac{L_{\alpha }^{n}f(z)\cdot {(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}}{z}\prec g(z),z\in U,$$

and this result is sharp. □

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

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

$${\left[{(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}\right]}^2+L_{\alpha }^{n}f(z)\cdot {(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }\mathrm{\prime }}\prec h(z),z\in U,$$

(2.12)

then

$$\frac{L_{\alpha }^{n}f(z)\cdot {(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}}{z}\prec q(z),z\in U,$$

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

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

Differentiating, we obtain ${[{(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}]}^2+L_{\alpha }^{n}f(z)\cdot {(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }\mathrm{\prime }}=p(z)+z{p}^{\mathrm{\prime }}(z)$, $z\in U$, and (2.12) becomes

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

Using Lemma 1.1, we have

$$p(z)\prec q(z),z\in U,\mathit{\text{i.e.}}\text{,}\frac{L_{\alpha }^{n}f(z)\cdot {(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}}{z}\prec q(z)=\frac{1}{z}{\int }_0^{z}h(t)dt,z\in U,$$

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

$${\left[{(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}\right]}^2+L_{\alpha }^{n}f(z)\cdot {(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }\mathrm{\prime }}\prec h(z),z\in U,$$

(2.13)

then

$$\frac{L_{\alpha }^{n}f(z)\cdot {(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}}{z}\prec q(z),z\in U,$$

where q is given by $q(z)=(2\beta -1)+2(1-\beta )\frac{ln(1+z)}{z}$, $z\in U$. 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 $p(z)=\frac{L_{\alpha }^{n}f(z)\cdot {(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}}{z}$, the differential subordination (2.13) becomes

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

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

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

 □

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

Let $f(z)=z+z^2$, $z\in U$. For $n=1$, $\alpha =2$, we obtain $L_2^{1}f(z)=-R^{1}f(z)+2S^{1}f(z)=-z{f}^{\mathrm{\prime }}(z)+2z{f}^{\mathrm{\prime }}(z)=z{f}^{\mathrm{\prime }}(z)=z+2z^2$, $z\in U$.

Then ${(L_2^{1}f(z))}^{\mathrm{\prime }}=1+4z$,

$$\begin{array}{c}\frac{L_2^{1}f(z)\cdot {(L_2^{1}f(z))}^{\mathrm{\prime }}}{z}=\frac{(z+2z^2)(1+4z)}{z}=8z^2+6z+1,\hfill \\ {\left[{(L_2^{1}f(z))}^{\mathrm{\prime }}\right]}^2+L_2^{1}f(z)\cdot {(L_2^{1}f(z))}^{\mathrm{\prime }\mathrm{\prime }}={(1+4z)}^2+(z+2z^2)\cdot 4=24z^2+12z+1.\hfill \end{array}$$

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

$$24z^2+12z+1\prec \frac{1-z}{1+z},z\in U,$$

induce

$$8z^2+6z+1\prec -1+\frac{2ln(1+z)}{z},z\in U.$$

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

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

$${\left(\frac{z}{L_{\alpha }^{n}f(z)}\right)}^{\delta }\frac{L_{\alpha }^{n+1}f(z)}{1-\delta }(\frac{{(L_{\alpha }^{n+1}f(z))}^{\mathrm{\prime }}}{L_{\alpha }^{n+1}f(z)}-\delta \frac{{(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}}{L_{\alpha }^{n}f(z)})\prec h(z),z\in U,$$

(2.14)

holds, then

$$\frac{L_{\alpha }^{n+1}f(z)}{z}\cdot {\left(\frac{z}{L_{\alpha }^{n}f(z)}\right)}^{\delta }\prec g(z),z\in U.$$

This result is sharp.

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

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

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

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

By using Lemma 1.2, we have

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

and this result is sharp. □

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

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

$${\left(\frac{z}{L_{\alpha }^{n}f(z)}\right)}^{\delta }\frac{L_{\alpha }^{n+1}f(z)}{1-\delta }(\frac{{(L_{\alpha }^{n+1}f(z))}^{\mathrm{\prime }}}{L_{\alpha }^{n+1}f(z)}-\delta \frac{{(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}}{L_{\alpha }^{n}f(z)})\prec h(z),z\in U,$$

(2.15)

then

$$\frac{L_{\alpha }^{n+1}f(z)}{z}\cdot {\left(\frac{z}{L_{\alpha }^{n}f(z)}\right)}^{\delta }\prec q(z),z\in U,$$

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

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

Differentiating, we obtain ${(\frac{z}{L_{\alpha }^{n}f(z)})}^{\delta }\frac{L_{\alpha }^{n+1}f(z)}{1-\delta }(\frac{{(L_{\alpha }^{n+1}f(z))}^{\mathrm{\prime }}}{L_{\alpha }^{n+1}f(z)}-\delta \frac{{(L_{\alpha }^{n}f(z))}^{\mathrm{\prime }}}{L_{\alpha }^{n}f(z)})=p(z)+\frac{1}{1-\delta }z{p}^{\mathrm{\prime }}(z)$, $z\in U$, and (2.15) becomes

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

Using Lemma 1.1, we have

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

and q is the best dominant. □