Subordination Properties of Certain Operators Concerning Fractional Integral and Libera Integral Operator

Abstract

The results contained in this paper are the result of a study regarding fractional calculus combined with the classical theory of differential subordination established for analytic complex valued functions. A new operator is introduced by applying the Libera integral operator and fractional integral of order $\lambda$ for analytic functions. Many subordination properties are obtained for this newly defined operator by using famous lemmas proved by important scientists concerned with geometric function theory, such as Eenigenburg, Hallenbeck, Miller, Mocanu, Nunokawa, Reade, Ruscheweyh and Suffridge. Results regarding strong starlikeness and convexity of order $\alpha$ are also discussed, and an example shows how the outcome of the research can be applied.

1. Introduction

Ever since the theory of differential subordination was initiated by Miller and Mocanu in the work published in 1978 [1] and 1981 [2], it was intensely used since it proves useful at re-obtaining known results in easier manners and also for providing interesting results when associated to studies involving analytic functions. A line of research which developed nicely in the context of differential subordination theory resulted after incorporating different types of operators into the study. Integral operators are an important tool when such investigations are considered as a recent survey paper shows [3]. The research started with the integral operator introduced by Alexander in 1915 [4]. A widely investigated integral operator is the Libera integral operator, introduced in 1965 [5]. Due to its properties of preserving starlikeness and convexity, it has been associated with many studies (see for example, references [6,7,8,9,10]) and still provides important new outcomes if combined with differential operators, such as in [11], with a confluent hypergeometric function, such as in [12], or with a generalized distribution, such as in [13]. Generalizations of the Libera operator are also considered for recent studies in papers, such as [14,15,16,17].

In the present investigation, the Libera integral operator is extended and combined with the fractional integral of order $\lambda$ for introducing a new fractional calculus operator. The idea was inspired by recent publications where the fractional integral is associated with the Mittag–Leffler confluent hypergeometric function [18,19,20], with the confluent hypergeometric function [21,22], with the Ruscheweyh and Sălăgean Operators [23], with the convolution product of the multiplier transformation and the Ruscheweyh derivative [24], with the convolution product of Sălăgean operator and Ruscheweyh derivative [25] or with other operators [26,27].

Consider the class of functions $f\left(z\right)$ of the form

$$f\left(z\right)=z+\sum _{k=2}^{\infty }{a}_k{z}^k$$

denoted by A and called analytic functions in the open unit disk $U=\{z\in \mathbb{C}:|z|<1\}$.

For $f\left(z\right)\in A$, Libera [5] introduced the integral operator $L_1\left(f\left(z\right)\right)$ defined as

$$\begin{array}{cc}\hfill L_1\left(f\left(z\right)\right)& {\displaystyle =\frac{2}{z}{\int }_0^{z}f\left(t\right)dt}\hfill \\ \hfill & =z+\sum _{k=2}^{\infty }\left({\displaystyle \frac{2}{k+1}}\right)a_k{z}^k.\hfill \end{array}$$

Consider the following extension for the operator $L_1\left(f\left(z\right)\right)$.

$$\begin{array}{cc}\hfill L_2\left(f\left(z\right)\right)& =L_1\left(L_1\left(f\left(z\right)\right)\right)\hfill \\ \hfill & =z+\sum _{k=2}^{\infty }{\left({\displaystyle \frac{2}{k+1}}\right)}^2{a}_k{z}^k\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill L_n\left(f\left(z\right)\right)& =L_1\left(L_{n-1}\left(f\left(z\right)\right)\right)\hfill \\ \hfill & =z+\sum _{k=2}^{\infty }{\left({\displaystyle \frac{2}{k+1}}\right)}^n{a}_k{z}^k,\hfill \end{array}$$

where $n\in \mathbb{N}=\{1,2,3,\dots \}$, and $L_0\left(f\left(z\right)\right)=f\left(z\right)$.

For $f\left(z\right)\in A$, the extension called fractional integral of order $\lambda$ is used in [28,29] as:

$${\displaystyle I_z^{\lambda }\left(f\left(z\right)\right)=\frac{1}{\Gamma \left(\lambda \right)}{\int }_0^z\frac{f\left(t\right)}{{(z-t)}^{1-\lambda }}dt(\lambda >0),}$$

where the multiplicity of ${(z-t)}^{\lambda -1}$ is removed by requiring $log(z-t)$ to be real when $z-t>0$, and $\Gamma \left(z\right)$ is the gamma function.

The following form in easily deduced:

$${\displaystyle I_z^{\lambda }\left(f\left(z\right)\right)=\frac{1}{\Gamma (2+\lambda )}{z}^{1+\lambda }+\sum _{k=2}^{\infty }\frac{k!}{\Gamma (k+1+\lambda )}{a}_k{z}^{k+\lambda }.}$$

Using $I_z^{\lambda }\left(f\left(z\right)\right)$, we consider

$$\begin{array}{cc}\hfill L_{\lambda }\left(f\left(z\right)\right)& {\displaystyle =\frac{\Gamma (2+\lambda )}{z^{\lambda }}{I}_z^{\lambda }\left(f\left(z\right)\right)}\hfill \\ \hfill & {\displaystyle =z+\sum _{k=2}^{\infty }\frac{k!\Gamma (2+\lambda )}{\Gamma (k+1+\lambda )}{a}_k{z}^k(\lambda >0).}\hfill \end{array}$$

(1)

It follows from the above that

$$L_0\left(f\left(z\right)\right)=\underset{\lambda \to 0}{lim}{L}_{\lambda }\left(f\left(z\right)\right)=f\left(z\right)$$

and

$$L_1\left(f\left(z\right)\right)=z+\sum _{k=2}^{\infty }\left({\displaystyle \frac{2}{k+1}}\right)a_k{z}^k.$$

Definition 1.

Using the operator $L_{\lambda }\left(f\left(z\right)\right)$ given by (1), we introduce

$$\begin{array}{cc}\hfill L_{n+\lambda }\left(f\left(z\right)\right)& =L_n\left(L_{\lambda }\left(f\left(z\right)\right)\right)\hfill \\ \hfill & {\displaystyle =z+\sum _{k=2}^{\infty }{\left({\displaystyle \frac{2}{k+1}}\right)}^n\frac{k!\Gamma (2+\lambda )}{\Gamma (k+1+\lambda )}{a}_k{z}^k}\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill L_{\lambda +n}\left(f\left(z\right)\right)& =L_{\lambda }\left(L_n\left(f\left(z\right)\right)\right)\hfill \\ \hfill & {\displaystyle =z+\sum _{k=2}^{\infty }{\left({\displaystyle \frac{2}{k+1}}\right)}^n\frac{k!\Gamma (2+\lambda )}{\Gamma (k+1+\lambda )}{a}_k{z}^k}\hfill \end{array}$$

for $n=0,1,2,\dots$ and $0<\lambda \le 1$. Considering the expressions above, we have:

$$L_{n+\lambda }\left(f\left(z\right)\right)=L_{\lambda +n}\left(f\left(z\right)\right).$$

For $f\left(z\right)\in A$, $f\left(z\right)$ is said to be subordinate to $g\left(z\right)$, written $f\left(z\right)\prec g\left(z\right)$, if there exists a function $w\left(z\right)$ analytic in U with $w\left(0\right)=0$ and $\left|w\right(z\left)\right|<1$$(z\in U)$, and such that $f\left(z\right)=g\left(w\right(z\left)\right)$. If $g\left(z\right)$ is univalent in U, then $f\left(z\right)\prec g\left(z\right)$ if and only if $f\left(0\right)=g\left(0\right)$ and $f\left(U\right)\subset g\left(U\right)$ ([30,31]).

We note that $f\left(z\right)\in A$ belongs to the class of starlike functions of order $\alpha$ in U if

$${\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\prec \frac{1+(1-2\alpha )z}{1-z}(z\in U)}$$

for $0\le \alpha <1$ and that $f\left(z\right)\in A$ belongs to the class of convex functions of order $\alpha$ in U if

$${\displaystyle 1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\prec \frac{1+(1-2\alpha )z}{1-z}(z\in U)}$$

for $0\le \alpha <1$.

In addition, the analytic function $p\left(z\right)$, $z\in U$, satisfies the condition

$${\displaystyle |argp\left(z\right)|<\frac{\pi }{2}\alpha (z\in U)}$$

for certain real values $\alpha >0$ if

$$p\left(z\right)\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\alpha }(z\in U).$$

In Section 2 of the paper, a series of properties are proved for the newly introduced operator $L_{\lambda }\left(f\left(z\right)\right)$ given by (1) considering the theory of differential subordination and a well-known lemma from Miller and Mocanu [30,32]. The study on operator $L_{\lambda }\left(f\left(z\right)\right)$ is continued in Section 3 with results obtained by using lemmas from Suffridge [33] and its improved form obtained by Hallenbeck and Ruscheweyh [34]. Results related to the Briot–Bouquet differential subordination involving the operator $L_{\lambda }\left(f\left(z\right)\right)$ are also obtained in Section 3 by using a lemma from Eenigenburg, Miller, Mocanu and Reade [35]. The study considering the operator $L_{\lambda }\left(f\left(z\right)\right)$ and known lemmas is concluded in Section 3 with two theorems that use a result proved by Nunokawa [36,37] for obtaining certain univalence conditions for the operator $L_{\lambda }\left(f\left(z\right)\right)$. The necessary lemmas cited above are listed in every section before each new result that is obtained as application. In Section 4, strong starlikeness and convexity of order $\alpha$ are investigated regarding the operator $L_{\lambda }\left(f\left(z\right)\right)$, and an example is also presented as an application for the new results.

2. Subordination Results Regarding ${\mathit{L}}_{\mathit{n}+\mathit{\lambda }}\left(\mathit{F}\left(\mathit{Z}\right)\right)$

To consider some properties of $L_{n+\lambda }\left(f\left(z\right)\right)$, the following result proved by Miller and Mocanu ([30,32]) (also from Jack [38]) will be considered in the study.

Lemma 1 ([30,32,38]).

Let $w\left(z\right)$ be analytic in U with $w\left(0\right)=0$. Then, if $\left|w\right(z\left)\right|$ attains its maximum value on the circle $\left|z\right|=r<1$ at a point $z_0\in U$, then we have

$$z_0{w}^{\prime }\left(z_0\right)=mw\left(z_0\right)$$

and

$$\mathrm{Re}\left({\displaystyle 1+\frac{z_0{w}^{″}\left(z_0\right)}{w^{\prime }\left(z_0\right)}}\right)\ge m,$$

where $m\ge 1$.

Using the lemma presented above, the following theorem can be stated and proved:

Theorem 1.

Consider the function $f\left(z\right)\in A$ satisfying the subordination

$${\displaystyle \frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}\prec \frac{\alpha (1+z)}{\alpha +(2-\alpha )z}(z\in U)}$$

(2)

for certain real values $\alpha >1$. The subordination (2) gives:

$${\displaystyle \left|{\displaystyle \frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}-\frac{\alpha }{2}}\right|<\frac{\alpha }{2}(z\in U).}$$

(3)

Proof. 

With condition (2), there exists an analytic function $w\left(z\right)$ satisfying the properties needed for the definition of subordination and

$${\displaystyle \frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}=\frac{\alpha (1+w(z\left)\right)}{\alpha +(2-\alpha )w\left(z\right)}(z\in U).}$$

(4)

Using relation (4) have that

$$\left|w\left(z\right)\right|=\left|{\displaystyle \frac{\alpha \left({\displaystyle \frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}-1}\right)}{{\displaystyle \alpha -(2-\alpha )\frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}}}}\right|<1(z\in U),$$

and that

$$2{\left|{\displaystyle \frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}}\right|}^2-\alpha \left({\displaystyle \frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}+\overline{\left({\displaystyle \frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}}\right)}}\right)<0$$

(5)

for $z\in U$. Hence, inequality (3) holds. □

Remark 1.

The result (3) in Theorem 1 shows us that

$$0<\mathrm{Re}\left({\displaystyle \frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}}\right)<\alpha (z\in U)$$

for $\alpha >1$.

Let us consider the analytic function $f\left(z\right)$ such that

$${\displaystyle L_{n-1+\lambda }\left(f\left(z\right)\right)=\frac{z(4+5z-2z^2)}{{(2-z)}^2}(z\in U).}$$

Then, we see that

$$\begin{array}{cc}\hfill L_{n+\lambda }\left(f\left(z\right)\right)& {\displaystyle =\frac{2}{z}{\int }_0^z{L}_{n-1+\lambda }\left(f\left(t\right)\right)dt}\hfill \\ \hfill & {\displaystyle =\frac{2}{z}{\int }_0^z\frac{t(4+5t-2t^2)}{{(2-t)}^2}dt}\hfill \\ \hfill & {\displaystyle =\frac{2z(1+z)}{2-z}.}\hfill \end{array}$$

(6)

The function obtained in (6) can be used in subordination (2) and satisfies the inequality (3) for $\alpha =4$.

For an analytic function $f\left(z\right)$, the following result can be proved.

Theorem 2.

If $f\left(z\right)\in A$ satisfies

$${\displaystyle \mathrm{Re}\left({\displaystyle \frac{L_{n-1+\lambda }\left(f\left(z\right)\right)}{L_{n+\lambda }\left(f\left(z\right)\right)}-1}\right)<\frac{1}{4(\alpha -1)}(z\in U)}$$

(7)

for $1<\alpha \le 2$ or

$${\displaystyle \mathrm{Re}\left({\displaystyle \frac{L_{n-1+\lambda }\left(f\left(z\right)\right)}{L_{n+\lambda }\left(f\left(z\right)\right)}-1}\right)<\frac{\alpha -1}{4}(z\in U)}$$

(8)

for $\alpha >2$, then

$${\displaystyle \left|{\displaystyle \frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}-\frac{\alpha }{2}}\right|<\frac{\alpha }{2}(z\in U).}$$

Proof. 

Consider an analytic function $w\left(z\right)$ that satisfies relation (4). We know that $w\left(0\right)=0$, and we obtain from (4) that

$${\displaystyle \frac{z{\left(L_{n+\lambda }\left(f\left(z\right)\right)\right)}^{\prime }}{L_{n+\lambda }\left(f\left(z\right)\right)}-1=\frac{z{w}^{\prime }\left(z\right)}{w\left(z\right)}\left({\displaystyle \frac{w\left(z\right)}{1-w\left(z\right)}-\frac{(2-\alpha )w\left(z\right)}{\alpha +(2-\alpha )w\left(z\right)}}\right).}$$

(9)

Since

$$z{\left(L_{n+\lambda }\left(f\left(z\right)\right)\right)}^{\prime }=2L_{n-1+\lambda }\left(f\left(z\right)\right)-L_{n+\lambda }\left(f\left(z\right)\right),$$

Equation (9) becomes

$${\displaystyle \frac{L_{n-1+\lambda }\left(f\left(z\right)\right)}{L_{n+\lambda }\left(f\left(z\right)\right)}-1=\frac{z{w}^{\prime }\left(z\right)}{2w\left(z\right)}\left({\displaystyle \frac{w\left(z\right)}{1-w\left(z\right)}-\frac{(2-\alpha )w\left(z\right)}{\alpha +(2-\alpha )w\left(z\right)}}\right).}$$

For the considered function $w\left(z\right)$, assume that there exists a point $z_0\in U$ such that

$$\underset{\left|z\right|\le |z_0|}{max}\left|w\left(z\right)\right|=|w\left(z_0\right)|=1.$$

In this situation, we write $w\left(z_0\right)=e^{i\theta }$$(0\le \theta <2\pi )$ and

$$z_0{w}^{\prime }\left(z_0\right)=kw\left(z_0\right)(k\ge 1).$$

Using the properties seen above, we have

$$\begin{array}{cc}\hfill \mathrm{Re}\left({\displaystyle \frac{L_{n-1+\lambda }\left(f\left(z_0\right)\right)}{L_{n+\lambda }\left(f\left(z_0\right)\right)}-1}\right)& {\displaystyle =\frac{k}{2}\mathrm{Re}\left({\displaystyle \frac{w\left(z_0\right)}{1-w\left(z_0\right)}-\frac{(2-\alpha )w\left(z_0\right)}{\alpha +(2-\alpha )w\left(z_0\right)}}\right)}\hfill \\ \hfill & {\displaystyle =\frac{k}{2}\left({\displaystyle \frac{1}{2}-\frac{(2-\alpha )(2-\alpha +\alpha cos\theta )}{{\alpha }^2+{(2-\alpha )}^2+2\alpha (2-\alpha )cos\theta }}\right).}\hfill \end{array}$$

Considering a function $g\left(t\right)$ given by

$${\displaystyle g\left(t\right)=\frac{2-\alpha +\alpha t}{{\alpha }^2+{(2-\alpha )}^2+2\alpha (2-\alpha )t}(t=cos\theta ),}$$

we have

$${\displaystyle g^{\prime }\left(t\right)=\frac{4\alpha (\alpha -1)}{{({\alpha }^2+{(2-\alpha )}^2+2\alpha (2-\alpha )t)}^2}>0.}$$

Since $g\left(t\right)$ is increasing for $t=cos\theta$, we obtain for $1<\alpha \le 2$ that

$${\displaystyle \mathrm{Re}\left({\displaystyle \frac{L_{n-1+\lambda }\left(f\left(z_0\right)\right)}{L_{n+\lambda }\left(f\left(z_0\right)\right)}-1}\right)\ge \frac{k}{4(\alpha -1)}\ge \frac{1}{4(\alpha -1)}}$$

(10)

and

$${\displaystyle \mathrm{Re}\left({\displaystyle \frac{L_{n-1+\lambda }\left(f\left(z_0\right)\right)}{L_{n+\lambda }\left(f\left(z_0\right)\right)}-1}\right)\ge \frac{(\alpha -1)k}{4}\ge \frac{\alpha -1}{4}(\alpha >2).}$$

(11)

Since (10) contradicts (7) and (11) contradicts (8), we say that there is no $w\left(z\right)$ such that $w\left(0\right)=0$ and $\left|w\right(z_0\left)\right|=1$ for $z_0\in U$. This implies that

$$\left|w\left(z\right)\right|=\left|{\displaystyle \frac{\alpha \left({\displaystyle \frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}-1}\right)}{{\displaystyle \alpha -(2-\alpha )\frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}}}}\right|<1(z\in U),$$

that is the inequality (5). □

Next, our result is

Theorem 3.

Consider the analytic function $f\left(z\right)$ satisfying the conditions

$${\displaystyle \mathrm{Re}\left({\displaystyle \frac{L_{n+\lambda }\left(f\left(z\right)\right)}{L_{n+1+\lambda }\left(f\left(z\right)\right)}-\frac{L_{n-1+\lambda }\left(f\left(z\right)\right)}{L_{n+\lambda }\left(f\left(z\right)\right)}-1}\right)<\frac{1}{4(\alpha -1)}(z\in U),}$$

for $1<\alpha \le 2$ or

$${\displaystyle \mathrm{Re}\left({\displaystyle \frac{L_{n+\lambda }\left(f\left(z\right)\right)}{L_{n+1+\lambda }\left(f\left(z\right)\right)}-\frac{L_{n-1+\lambda }\left(f\left(z\right)\right)}{L_{n+\lambda }\left(f\left(z\right)\right)}-1}\right)<\frac{\alpha -1}{4}(z\in U),}$$

for $\alpha >2$.

Then,

$${\displaystyle \left|{\displaystyle \frac{L_{n+1+\lambda }\left(f\left(z\right)\right)}{L_{n+\lambda }\left(f\left(z\right)\right)}-\frac{\alpha }{2}}\right|<\frac{\alpha }{2}(z\in U).}$$

Proof. 

Consider a function $w\left(z\right)$ satisfying

$${\displaystyle \frac{L_{n+1+\lambda }\left(f\left(z\right)\right)}{L_{n+\gamma }\left(f\left(z\right)\right)}=\frac{\alpha (1+w(z\left)\right)}{\alpha +(2-\alpha )w\left(z\right)}(z\in U).}$$

This shows that $w\left(0\right)=0$.

Using

$$z{\left(L_{n+1+\lambda }\left(f\left(z\right)\right)\right)}^{\prime }=2L_{n+\lambda }\left(f\left(z\right)\right)-L_{n+1+\lambda }\left(f\left(z\right)\right)$$

and

$$z{\left(L_{n+\lambda }\left(f\left(z\right)\right)\right)}^{\prime }=2L_{n-1+\lambda }\left(f\left(z\right)\right)-L_{n+\lambda }\left(f\left(z\right)\right),$$

we obtain that

$${\displaystyle \frac{L_{n+\lambda }\left(f\left(z\right)\right)}{L_{n+1+\lambda }\left(f\left(z\right)\right)}-\frac{L_{n-1+\lambda }\left(f\left(z\right)\right)}{L_{n+\lambda }\left(f\left(z\right)\right)}-1=\frac{z{w}^{\prime }\left(z\right)}{2w\left(z\right)}\left({\displaystyle \frac{w\left(z\right)}{1-w\left(z\right)}-\frac{(2-\alpha )w\left(z\right)}{\alpha +(2-\alpha )w\left(z\right)}}\right).}$$

From this point on, the proof of this theorem is completed by following the same steps as for the proof of Theorem 2. □

3. Applications of Subordinations by Suffridge

We first introduce the following lemma proved by Suffridge [27].

Lemma 2 ([27]).

If a function $p\left(z\right)$ is analytic in U with $p\left(0\right)=1$ and satisfies

$$z{p}^{\prime }\left(z\right)\prec h\left(z\right)(z\in U)$$

for some starlike function $h\left(z\right)$, then

$${\displaystyle p\left(z\right)\prec {\int }_0^z\frac{h\left(t\right)}{t}(z\in U).}$$

Applying the above lemma, we have

Theorem 4.

Consider the analytic function $f\left(z\right)$ satisfying the following subordination

$${\displaystyle \frac{L_{n-1+\lambda }\left(f\left(z\right)\right)-L_{n+\lambda }\left(f\left(z\right)\right)}{z}\prec \frac{1+(1-2\alpha )z}{2(1-z)}(z\in U),}$$

for certain real values α$(0\le \alpha <1)$.

Then

$${\displaystyle \frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}\prec log\left({\displaystyle \frac{\sqrt{z}}{{(1-z)}^{1-\alpha }}}\right)(z\in U).}$$

Proof. 

Consider the analytic function $p\left(z\right)$ with $p\left(0\right)=1$ given by:

$${\displaystyle p\left(z\right)=\frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}.}$$

In addition, consider the starlike function of order $\alpha$$h\left(z\right)$ given by

$${\displaystyle h\left(z\right)=\frac{1+(1-2\alpha )z}{1-z}(z\in U),}$$

for $0\le \alpha <1$.

Since

$$\begin{array}{cc}\hfill z{p}^{\prime }\left(z\right)& {\displaystyle =\frac{z{\left(L_{n+\lambda }\left(f\left(z\right)\right)\right)}^{\prime }-L_{n+\lambda }\left(f\left(z\right)\right)}{z}}\hfill \\ \hfill & {\displaystyle =\frac{2(L_{n-1+\lambda }\left(f\left(z\right)\right)-L_{n+\gamma }\left(f\left(z\right)\right))}{z}}\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill {\displaystyle {\int }_0^z\frac{h\left(t\right)}{t}dt}& ={\int }_0^z\left({\displaystyle \frac{1}{t}-\frac{2(1-\alpha )}{1-t}}\right)dt\hfill \\ \hfill & =log\left({\displaystyle \frac{z}{{(1-z)}^{2(1-\alpha )}}}\right),\hfill \end{array}$$

by applying Lemma 2, we obtain that

$${\displaystyle \frac{L_{n-1+\lambda }\left(f\left(z\right)\right)-L_{n+\lambda }\left(f\left(z\right)\right)}{z}\prec \frac{1}{2}log\left({\displaystyle \frac{z}{{(1-z)}^{2(1-\alpha )}}}\right)(z\in U).}$$

Hence, the proof is completed. □

Taking ${\displaystyle \alpha =\frac{1}{2}}$ in Theorem 4, the following corollary emerges:

Corollary 1.

If $f\left(z\right)\in A$ satisfies

$${\displaystyle \frac{L_{n-1+\lambda }\left(f\left(z\right)\right)-L_{n+\lambda }\left(f\left(z\right)\right)}{z}\prec \frac{1}{1-z}(z\in U),}$$

then

$${\displaystyle \frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}\prec \frac{1}{2}log\left({\displaystyle \frac{z}{1-z}}\right)(z\in U).}$$

Hallenbeck and Ruscheweyh [28] obtained the following form for Lemma 2 given by Suffridge:

Lemma 3 ([28]).

If a function $p\left(z\right)$ is analytic in U with $p\left(0\right)=1$ and satisfies

$$p\left(z\right)+z{p}^{\prime }\left(z\right)\prec h\left(z\right)(z\in U)$$

for some convex function $h\left(z\right)$, then

$${\displaystyle p\left(z\right)\prec \frac{1}{z}{\int }_0^{z}h\left(t\right)dt(z\in U).}$$

Now, we prove the following result.

Theorem 5.

Consider the analytic function $f\left(z\right)$ satisfying the subordination

$${\displaystyle 2\frac{L_{n-1+\lambda }\left(f\left(z\right)\right)}{z}-\frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}\prec log\left({\displaystyle \frac{z}{{(1-z)}^{2(1-\alpha )}}}\right)(z\in U),}$$

for certain real values of $\alpha (0\le \alpha <1)$.

Then,

$${\displaystyle \frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}\prec log\left({\displaystyle \frac{z}{{(1-z)}^{2(1-\alpha )}}}\right)+\frac{2(1-\alpha )}{z}log(1-z)+(1-2\alpha )(z\in U).}$$

Proof. 

Consider the analytic function $p\left(z\right)$, $z\in U$, with $p\left(0\right)=1$, given by

$${\displaystyle p\left(z\right)=\frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}.}$$

Using it, we can write:

$${\displaystyle p\left(z\right)+z{p}^{\prime }\left(z\right)=2\frac{L_{n-1+\lambda }\left(f\left(z\right)\right)}{z}-\frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}.}$$

Further, we know that a function $h\left(z\right)$ given by

$$h\left(z\right)=log\left({\displaystyle \frac{z}{{(1-z)}^{2(1-\alpha )}}}\right)(z\in U)$$

satisfies

$${\displaystyle z{h}^{\prime }\left(z\right)=\frac{1+(1-2\alpha )z}{1-z}.}$$

Thus, $h\left(z\right)$ is convex in U because $z{h}^{\prime }\left(z\right)$ is starlike of order $\alpha$ in U. Applying Lemma 3, we obtain

$$\begin{array}{cc}\hfill {\displaystyle \frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}}& {\displaystyle \prec \frac{1}{z}{\int }_0^z\left(log\left({\displaystyle \frac{t}{{(1-t)}^{2(1-\alpha )}}}\right)\right)dt}\hfill \\ \hfill & {\displaystyle =\frac{1}{z}{\int }_0^z(logt-2(1-\alpha )log(1-t))dt}\hfill \\ \hfill & {\displaystyle =log\left({\displaystyle \frac{z}{{(1-z)}^{2(1-\alpha )}}}\right)+\frac{2(1-\alpha )}{z}log(1-z)+(1-2\alpha )(z\in U).}\hfill \end{array}$$

 □

Choosing ${\displaystyle \alpha =\frac{1}{2}}$ in Theorem 5, we obtain the following corollary.

Corollary 2.

Consider the analytic function $f\left(z\right)$ satisfying the following subordination:

$${\displaystyle 2\frac{L_{n-1+\lambda }\left(f\left(z\right)\right)}{z}-\frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}\prec log\left({\displaystyle \frac{z}{1-z}}\right)(z\in U)}$$

Then,

$${\displaystyle \frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}\prec log\left({\displaystyle \frac{z}{1-z}}\right)+\frac{1}{z}log(1-z)(z\in U).}$$

Theorem 6.

Consider the analytic function $f\left(z\right)$ satisfying the following subordination:

$${\displaystyle 2\frac{L_{n-1+\lambda }\left(f\left(z\right)\right)}{z}-\frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}\prec \frac{1+z}{1-z}(z\in U).}$$

Then,

$${\displaystyle \frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}\prec \frac{2}{z}log\left({\displaystyle \frac{1}{1-z}}\right)-1(z\in U).}$$

(12)

Proof. 

Letting

$${\displaystyle p\left(z\right)=\frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z} \mathrm{and} h\left(z\right)=\frac{1+z}{1-z},}$$

we have that $p\left(z\right)$ is analytic in U with $p\left(0\right)=1$, and $h\left(z\right)$ is convex in U. Since

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{z}{\int }_0^{z}h\left(t\right)dt}& {\displaystyle =\frac{1}{z}{\int }_0^z\left({\displaystyle \frac{1+t}{1-t}}\right)dt}\hfill \\ \hfill & {\displaystyle =\frac{2}{z}log\left({\displaystyle \frac{1}{1-z}}\right)-1,}\hfill \end{array}$$

we have the subordination (12). □

Next, the lemma given below, proved by Eenigenburg, Miller, Mocanu and Reade [29], is used for obtaining a new result.

Lemma 4 ([29]).

Let $h\left(z\right)$ be convex in U with $\mathrm{Re}\left(\beta h\right(z)+\gamma )>0$$(\beta \ne 0)$. If $p\left(z\right)$ is analytic in U with $p\left(0\right)=h\left(0\right)$, then the subordination

$${\displaystyle p\left(z\right)+\frac{z{p}^{\prime }\left(z\right)}{\beta p\left(z\right)+\gamma }\prec h\left(z\right)(z\in U)}$$

satisfies

$$p\left(z\right)\prec h\left(z\right)(z\in U).$$

With this lemma, we have

Theorem 7.

Consider the analytic function $f\left(z\right)$ satisfying the following subordination:

$${\displaystyle \frac{L_{n+1+\lambda }\left(f\left(z\right)\right)-2L_{n-1+\lambda }\left(f\left(z\right)\right)}{L_{n+\lambda }\left(f\left(z\right)\right)}+\frac{2L_{n+\lambda }\left(f\left(z\right)\right)}{L_{n+1+\lambda }\left(f\left(z\right)\right)}\prec \frac{1+(1-2\alpha )z}{1-z}(z\in U),}$$

for $0\le \alpha <1$.

Then,

$${\displaystyle \frac{L_{n+1+\lambda }\left(f\left(z\right)\right)}{L_{n+\lambda }\left(f\left(z\right)\right)}\prec \frac{1+(1-2\alpha )z}{1-z}(z\in U).}$$

Proof. 

Consider the analytic function $p\left(z\right)$, $z\in U$, with $p\left(0\right)=1$, given by

$${\displaystyle p\left(z\right)=\frac{L_{n+1+\lambda }\left(f\left(z\right)\right)}{L_{n+\lambda }\left(f\left(z\right)\right)}.}$$

In addition, consider the convex function of order $\alpha$ given by

$${\displaystyle h\left(z\right)=\frac{1+(1-2\alpha )z}{1-z},0\le \alpha <1,}$$

$h\left(0\right)=1$.

Taking $\beta =1$ and $\gamma =0$ in Lemma 4, we say that

$${\displaystyle p\left(z\right)+\frac{z{p}^{\prime }\left(z\right)}{p\left(z\right)}\prec \frac{1+(1-2\alpha )z}{1-z}(z\in U)}$$

implies

$${\displaystyle p\left(z\right)\prec \frac{1+(1-2\alpha )z}{1-z}(z\in U).}$$

Since

$${\displaystyle \frac{z{p}^{\prime }\left(z\right)}{p\left(z\right)}=2\left({\displaystyle \frac{L_{n+\lambda }\left(f\left(z\right)\right)}{L_{n+1+\lambda }\left(f\left(z\right)\right)}-\frac{L_{n-1+\lambda }\left(f\left(z\right)\right)}{L_{n+\lambda }\left(f\left(z\right)\right)}}\right),}$$

we prove the theorem with Lemma 4. □

Next, we consider the following lemma proved by Nunokawa ([30,31]).

Lemma 5 ([30,31]).

Let a function $p\left(z\right)$ be analytic in U with $p\left(0\right)=1$. If there exists a point $z_0$$\left(\right|z_0|<1)$ such that

$${\displaystyle |arg\left(p\left(z\right)\right)|<\frac{\pi }{2}\beta \left(\right|z|<|z_0\left|\right)}$$

and

$${\displaystyle |arg\left(p\left(z_0\right)\right)|=\frac{\pi }{2}\beta }$$

for some real $\beta >0$, then

$${\displaystyle \frac{z_0{p}^{\prime }\left(z_0\right)}{p\left(z_0\right)}=\frac{2ikarg\left(p\left(z_0\right)\right)}{\pi }}$$

for some real k such that

$${\displaystyle k\ge \frac{1}{2}\left({\displaystyle a+\frac{1}{a}}\right)>1,}$$

where

$${\left(p\left(z_0\right)\right)}^{\frac{1}{\beta }}=\pm ia(a>0).$$

Now, we derive

Theorem 8.

Let $f\left(z\right)\in A$ and

$${\displaystyle F\left(z\right)=\frac{L_{n+\lambda }\left(f\left(z\right)\right)-\alpha z}{(1-\alpha )z}+2\frac{L_{n-1+\lambda }\left(f\left(z\right)\right)-\alpha z}{L_{n+\lambda }\left(f\left(z\right)\right)-\alpha z}-2}$$

(13)

for $0\le \alpha <1$. If $f\left(z\right)$ satisfies

$${\displaystyle F{\left(z\right)}^2-1\prec \frac{16z}{{(1-z)}^2}(z\in U),}$$

(14)

then,

$$\mathrm{Re}\left({\displaystyle \frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}}\right)>\alpha (z\in U).$$

Proof. 

Consider the analytic function $p\left(z\right)$, $z\in U$, with $p\left(0\right)=1$, given by:

$${\displaystyle p\left(z\right)=\frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}.}$$

For such $p\left(z\right)$, assume that there exists a point $z_0$$\left(\right|z_0|<1)$ such that

$$\mathrm{Re}\left({\displaystyle \frac{p\left(z\right)-\alpha }{1-\alpha }}\right)>0\left(\right|z|<|z_0|<1)$$

and

$$\mathrm{Re}\left({\displaystyle \frac{p\left(z_0\right)-\alpha }{1-\alpha }}\right)=0.$$

If

$${\displaystyle \frac{p\left(z_0\right)-\alpha }{1-\alpha }\ne 0,}$$

Lemma 5 gives us that

$$\begin{array}{cc}\hfill {\displaystyle \frac{z_0{p}^{\prime }\left(z_0\right)}{p\left(z_0\right)-\alpha }}& {\displaystyle =\frac{2ik}{\pi }arg\left({\displaystyle \frac{p\left(z_0\right)-\alpha }{1-\alpha }}\right)}\hfill \\ \hfill & {\displaystyle =\frac{2ik}{\pi }arg(p\left(z_0\right)-\alpha )}\hfill \end{array}$$

for some real k such that ${\displaystyle k\ge \frac{1}{2}\left({\displaystyle a+\frac{1}{a}}\right)>1}$ with

$${\left({\displaystyle \frac{p\left(z_0\right)-\alpha }{1-\alpha }}\right)}^{\frac{1}{\beta }}=\pm ia(a>0).$$

It follows from the above that

$$\begin{array}{cc}\hfill \left\{{\displaystyle \frac{p\left(z_0\right)-\alpha }{1-\alpha }+\frac{z_0{p}^{\prime }\left(z_0\right)}{p\left(z_0\right)-\alpha }}\right\}-1& =F{\left(z_0\right)}^2-1\hfill \\ \hfill & ={(\pm ia\pm ik)}^2-1\hfill \\ \hfill & \le -{\left({\displaystyle a+\frac{a^2+1}{2a}}\right)}^2-1.\hfill \end{array}$$

Let us consider a function $h\left(a\right)$ given by

$${\displaystyle h\left(a\right)=a+\frac{a^2+1}{2a}(a>0).}$$

Then, $h\left(a\right)$ satisfies

$$h\left(a\right)\ge h\left(\sqrt{{\displaystyle \frac{1}{3}}}\right)=\sqrt{3}.$$

This gives us that

$${\left\{{\displaystyle \frac{p\left(z_0\right)-\alpha }{1-\alpha }+\frac{z_0{p}^{\prime }\left(z_0\right)}{p\left(z_0\right)-\alpha }}\right\}}^2-1=F{\left(z_0\right)}^2-1\le -4.$$

Here, we define a function $g\left(z\right)$ by

$${\displaystyle g\left(z\right)=\frac{16z}{{(1-z)}^2}(z\in U).}$$

Then, $g\left(z\right)$ maps U onto the domain with the slit $(-\infty ,-4)$. This contradicts our condition (14).

Having the contradiction, we conclude that $p\left(z\right)$ satisfies the condition

$$\mathrm{Re}\left({\displaystyle \frac{p\left(z\right)-\alpha }{1-\alpha }}\right)=\mathrm{Re}\left({\displaystyle \frac{{\displaystyle \frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}-\alpha }}{1-\alpha }}\right)>0,$$

for all $z\in U$. Hence, the proof of the theorem is completed. □

Next, our theorem is

Theorem 9.

Consider a function $F\left(z\right)$ given by (13) where $f\left(z\right)$ is analytic in U and $0\le \alpha <1$. If $F\left(z\right)$ satisfies

$${\displaystyle F\left(z\right)\prec \frac{1+z}{1-z}(z\in U),}$$

(15)

then

$$\mathrm{Re}\left({\displaystyle \frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}}\right)>\alpha (z\in U).$$

Proof. 

Consider the analytic function $p\left(z\right)$ given by

$${\displaystyle p\left(z\right)=\frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}.}$$

Then, there exists a point $z_0$$\left(\right|z_0|<1)$ such that

$$\mathrm{Re}\left({\displaystyle \frac{p\left(z\right)-\alpha }{1-\alpha }}\right)>0\left(\right|z|<|z_0|<1)$$

(16)

and

$$\mathrm{Re}\left({\displaystyle \frac{p\left(z_0\right)-\alpha }{1-\alpha }}\right)=0.$$

(17)

If

$${\displaystyle \frac{p\left(z_0\right)-\alpha }{1-\alpha }=0,}$$

by Lemma 5, we have

$${\displaystyle \frac{z_0{p}^{\prime }\left(z_0\right)}{p\left(z_0\right)-\alpha }=\frac{2ik}{\pi }arg(p\left(z_0\right)-\alpha )}$$

for some real ${\displaystyle k\ge \frac{1}{2}\left({\displaystyle a+\frac{1}{a}}\right)>1}$ with

$${\displaystyle \frac{p\left(z_0\right)-\alpha }{1-\alpha }=\pm ia(a>0).}$$

With the properties obtained so far, we can write

$${\displaystyle \frac{p\left(z_0\right)-\alpha }{1-\alpha }+\frac{z_0{p}^{\prime }\left(z_0\right)}{p\left(z_0\right)-\alpha }=F\left(z_0\right)=\pm i(a+k).}$$

Since

$$\mathrm{Re}\left({\displaystyle \frac{1+z}{1-z}}\right)>0(z\in U),$$

we say that

$${\displaystyle F\left(z\right)\nprec \frac{1+z}{1-z}(z\in U).}$$

This means that there is no $z_0$$\left(\right|z_0|<1)$ such that (16) and (17) are satisfied. Hence, we obtain the stated conclusion of the theorem.

$$\mathrm{Re}(p\left(z\right)-\alpha )=\mathrm{Re}\left({\displaystyle \frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}-\alpha }\right)>0(z\in U).$$

 □

Remark 2.

Considering $f\left(z\right)$ an analytic function in U given by

$${\displaystyle \frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}=\frac{1}{1-z}}$$

and $\alpha =0$ in Theorem 9, we have that

$${\displaystyle F\left(z\right)=\frac{1+z}{1-z}(z\in U).}$$

Therefore, $f\left(z\right)$ satisfies the subordination (15) for $\alpha =0$. For such $f\left(z\right)$, we know that

$${\displaystyle \mathrm{Re}\left({\displaystyle \frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}}\right)>\frac{1}{2}>0(z\in U).}$$

4. Results Regarding Strong Properties of Order $\mathbf{\alpha }$

Let $f\left(z\right)\in A$ and $L_{n+\lambda }\left(f\left(z\right)\right)$ be defined by (1) for $n=0,1,2,\dots$ and $0\le \lambda \le 1$. For $f\left(z\right)\in A$ satisfying

$$\mathrm{Re}\left({\displaystyle \frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}}\right)>\alpha (z\in U),$$

$f\left(z\right)$ is said to be strongly of order $\alpha$ in U if $f\left(z\right)$ satisfies

$${\displaystyle \left|arg\left({\displaystyle \frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}}\right)\right|<\frac{\pi }{2}\alpha (z\in U),}$$

where $0\le \alpha <1$.

If $f\left(z\right)\in A$ satisfies

$${\displaystyle \left|arg\left({\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}\right)\right|<\frac{\pi }{2}\alpha (z\in U)}$$

for $0\le \alpha <1$, then $f\left(z\right)$ is said to be strongly starlike of order $\alpha$ in U. In addition, if $f\left(z\right)\in A$ satisfies

$${\displaystyle \left|arg\left({\displaystyle 1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}}\right)\right|<\frac{\pi }{2}\alpha (z\in U)}$$

for $0\le \alpha <1$, then we say that $f\left(z\right)$ is strongly convex of order $\alpha$ in U.

Let us consider a function $w\left(z\right)$ defined by

$$w\left(z\right)={\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\alpha }(z\in U)$$

for $0\le \alpha <1$, then we see that

$${\displaystyle argw\left(z\right)=\alpha arg\left({\displaystyle \frac{1+z}{1-z}}\right)=\frac{\pi }{2}\alpha (z\in U).}$$

Thus, a function $f\left(z\right)$ given by

$$f\left(z\right)=exp\left({\int }_0^z{\left({\displaystyle \frac{1+t}{1-t}}\right)}^{\alpha }dt\right)$$

is strongly starlike of order $\alpha$ in U and a function $f\left(z\right)\in A$ given by

$${\displaystyle f^{\prime }\left(z\right)=\frac{1}{z}exp\left({\int }_0^z{\left({\displaystyle \frac{1+t}{1-t}}\right)}^{\alpha }dt\right)}$$

is strongly convex of order $\alpha$ in U.

Now, we derive

Theorem 10.

If $f\left(z\right)\in A$ satisfies

$${\displaystyle \left|{\displaystyle \frac{L_{n-1+\lambda }\left(f\left(z\right)\right)}{L_{n+\lambda }\left(f\left(z\right)\right)}-1}\right|<\frac{\alpha }{4}\mathrm{Re}\left({\displaystyle \frac{1+\beta z}{1-z}}\right)(z\in U)}$$

for some real α$(0\le \alpha <1)$ and some real β $(\beta \ne -1)$, then

$${\displaystyle \left|arg\left({\displaystyle \frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}}\right)\right|<\frac{\pi }{2}\alpha (z\in U).}$$

Proof. 

Define a function $p\left(z\right)$ by

$${\displaystyle p\left(z\right)=\frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}.}$$

Then, $p\left(z\right)$ is analytic in U, and $p\left(0\right)=1$. This function $p\left(z\right)$ satisfies

$${\displaystyle \frac{z{p}^{\prime }\left(z\right)}{p\left(z\right)}=2\left({\displaystyle \frac{L_{n-1+\lambda }\left(f\left(z\right)\right)}{L_{n+\lambda }\left(f\left(z\right)\right)}-1}\right).}$$

It follows from the above that

$$\begin{array}{cc}\hfill \left|arg\left({\displaystyle \frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}}\right)\right|& =|arg\left(p\left(z\right)\right)|=\left|\mathrm{Im}(log\left(p\left(z\right)\right))\right|=\left|\mathrm{Im}{\int }_0^z{(log\left(p\left(t\right)\right))}^{\prime }dt\right|\hfill \\ \hfill & =\left|{\displaystyle \mathrm{Im}{\int }_0^z\frac{p^{\prime }\left(t\right)}{p\left(t\right)}dt}\right|=\left|{\displaystyle \mathrm{Im}{\int }_0^z\frac{p^{\prime }\left(\rho e^{i\theta }\right)}{p\left(\rho e^{i\theta }\right)}{e}^{i\theta }d\rho }\right|\hfill \\ \hfill & \le {\int }_0^r\left|\mathrm{Im}\left({\displaystyle \frac{p^{\prime }\left(\rho e^{i\theta }\right)}{p\left(\rho e^{i\theta }\right)}{e}^{i\theta }}\right)\right|d\rho \le {\int }_{-r}^r\left|{\displaystyle \frac{p^{\prime }\left(\rho e^{i\theta }\right)}{p\left(\rho e^{i\theta }\right)}}\right|d\rho \hfill \\ \hfill & {\displaystyle \le \frac{r}{2}{\int }_0^{2\pi }\left|{\displaystyle \frac{p^{\prime }\left(r{e}^{i\theta }\right)}{p\left(r{e}^{i\theta }\right)}}\right|d\theta =\frac{1}{2}{\int }_0^{2\pi }\left|{\displaystyle \frac{r{e}^{i\theta }{p}^{\prime }\left(r{e}^{i\theta }\right)}{p\left(r{e}^{i\theta }\right)}}\right|d\theta }\hfill \\ \hfill & {\displaystyle ={\int }_0^{2\pi }\left|{\displaystyle \frac{L_{n-1+\lambda }\left(f\left(r{e}^{i\theta }\right)\right)}{L_{n+\lambda }\left(f\left(r{e}^{i\theta }\right)\right)}-1}\right|dt<\frac{\alpha }{4}{\int }_0^{2\pi }\mathrm{Re}\left({\displaystyle \frac{1+\beta r{e}^{i\theta }}{1-r{e}^{i\theta }}}\right)d\theta }\hfill \\ \hfill & {\displaystyle =\frac{\alpha }{4}{\int }_0^{2\pi }\left\{{\displaystyle \frac{1-\beta }{2}+\left({\displaystyle \frac{1+\beta }{2}}\right)\frac{1-r^2}{1+r^2-2rcos\theta }}\right\}d\theta =\frac{\pi }{2}\alpha ,}\hfill \end{array}$$

because by Poisson integral

$${\displaystyle \frac{1}{2\pi }{\int }_0^{2\pi }\frac{1-r^2}{1+r^2-2rcos\theta }d\theta =1.}$$

This completes the proof of the theorem. □

Example 1.

Consider a function $f\left(z\right)\in A$ given by

$$L_{n+\lambda }\left(f\left(z\right)\right)=z{\left({\displaystyle \frac{2}{2-z}}\right)}^{3\alpha }(z\in U),$$

with $0\le \alpha <1$. Note that a function

$${\displaystyle w\left(z\right)=\frac{2}{2-z}}$$

satisfies

$${\displaystyle \left|{\displaystyle w\left(z\right)-\frac{4}{3}}\right|<\frac{2}{3}(z\in U)}$$

and

$${\displaystyle |argw\left(z\right)|<\frac{\pi }{6}(z\in U).}$$

This gives us that

$${\displaystyle \left|arg\left({\displaystyle \frac{L_{n+\lambda }\left(f\left(z\right)\right)}{z}}\right)\right|<\frac{\pi }{2}\alpha (z\in U).}$$

For such $f\left(z\right)$, we have

$${\displaystyle \left|{\displaystyle \frac{L_{n-1+\lambda }\left(f\left(z\right)\right)}{L_{n+\lambda }\left(f\left(z\right)\right)}-1}\right|=\frac{3}{2}\alpha \left|{\displaystyle \frac{z}{2-z}}\right|<\frac{3}{2}\alpha (z\in U).}$$

Thus, if we consider a real β such that $\beta \le -11$, then $f\left(z\right)$ satisfies

$${\displaystyle \left|{\displaystyle \frac{L_{n-1+\lambda }\left(f\left(z\right)\right)}{L_{n+\lambda }\left(f\left(z\right)\right)}-1}\right|<\frac{3}{2}\alpha \le \frac{\alpha (1-\beta )}{8}<\frac{\alpha }{4}\mathrm{Re}\left({\displaystyle \frac{1+\beta z}{1-z}}\right),}$$

for $z\in U$.

5. Conclusions

The outcome of this paper falls within the research topic which concerns incorporating fractional calculus in geometric function theory by defining new fractional operators and conducting studies involving the theory of differential subordination. The operator used for the investigation denoted by $L_{\lambda }\left(f\left(z\right)\right)$ is introduced in Definition 1 using fractional integral of order $\lambda$ defined in [28,29] and the Libera integral operator [5]. The necessary known definitions regarding the analytic functions are shown in the Introduction. Section 2 contains three theorems that show the results of the study conducted on the operator $L_{\lambda }\left(f\left(z\right)\right)$ by applying a famous lemma from Miller and Mocanu [30,32] which is presented at the beginning of this section. Section 3 starts with recalling the lemma from Suffridge [33] which is used for obtaining the new results regarding the operator $L_{\lambda }\left(f\left(z\right)\right)$ contained in Theorem 4 and Corollary 1. This lemma was modified by Hallenbeck and Ruscheweyh [34]. Their resulting lemma is first listed, and then Theorems 5 and 6 and Corollary 2 present the new results obtained by applying it to the operator $L_{\lambda }\left(f\left(z\right)\right)$. A lemma from Eenigenburg, Miller, Mocanu and Reade [35] is next stated and used for obtaining the new result involving the operator $L_{\lambda }\left(f\left(z\right)\right)$ presented in Theorem 7. A lemma proved by Nunokawa [36,37] is next listed and applied to the operator $L_{\lambda }\left(f\left(z\right)\right)$ for the new outcome presented in Theorems 8 and 9. In Section 4, the basic definition regarding strong starlikeness and strong convexity of order $\alpha$ are recalled, and a new result concerning the strong starlikeness of order $\alpha$ of the operator $L_{\lambda }\left(f\left(z\right)\right)$ is proved. An example is also provided in order to show a certain application of the theoretical result presented in Theorem 10.

As future uses of the results presented here, the operator $L_{\lambda }\left(f\left(z\right)\right)$ given by (1) can be applied for defining new subclasses of analytic functions with certain geometric properties given by the characteristics of this operator already proven in this paper. The classes could be further investigated considering the strong starlikeness of order $\alpha$ of the operator $L_{\lambda }\left(f\left(z\right)\right)$ having as inspiration recent studies such as [39].