Applications of the Atangana–Baleanu Fractional Integral Operator

Abstract

Applications of the Atangana–Baleanu fractional integral were considered in recent studies related to geometric function theory to obtain interesting differential subordinations. Additionally, the multiplier transformation was used in many studies, providing elegant results. In this paper, a new operator is defined by combining those two prolific functions. The newly defined operator is applied for introducing a new subclass of analytic functions, which is investigated concerning certain properties, such as coefficient estimates, distortion theorems, closure theorems, neighborhoods and radii of starlikeness, convexity and close-to-convexity. This class may have symmetric or asymmetric properties. The results could prove interesting due to the new applications of the Atangana–Baleanu fractional integral and of the multiplier transformation. Additionally, the univalence properties of the new subclass of functions could inspire researchers to conduct further investigations related to this newly defined class.

1. Introduction

Fractional calculus has many applications in diverse fields of research. The papers [1,2] discuss the history of fractional calculus and provide references to its many applications in science and engineering.

Owa [3] and Owa and Srivastava [4] applied a fractional integral for a function which gave new possibilities for studying properties of functions and for defining new operators. A fractional integral was considered on a confluent hypergeometric function in recently published papers (see [5]) and on Ruscheweyh and Sălăgean operators in [6]. Applications of fractional derivatives with Mittag–Leffler kernels were considered in [7,8] and with non-local and non-singular kernels in [9].

Atangana and Baleanu [10] used the Riemann–Liouville fractional integral for introducing a new fractional integral studied by many researchers in recent years. The Atangana–Baleanu fractional integral of Bessel functions was used in studies [11,12]. Nice results were recently obtained regarding Ostrowski-type integral inequalities [13] and Hermite–Hadamard-type inequalities [14] involving an Atangana–Baleanu fractional integral operator. The definition given by Atangana–Baleanu can be extended to complex values of the order of differentiation $\nu$ by using analytic continuation.

Multiplier transformation has also been used for recent studies, as can be seen in [15,16].

Inspired by the nice results seen in the papers published considering the Atangana—-Baleanu fractional integral and multiplier transformation separately, we have decided to merge them and to define a new operator, which will be given below. This operator is used to introduce a new subclass of analytic functions, since introducing and studying new classes of univalent functions generates very interesting results, and we can remind only a few very recent results, such as new subclasses for bi-univalent functions [17,18] and classes of functions introduced using operators [19,20].

The class of analytic function is denoted by $\mathcal{H}\left(U\right)$ in $U=\{z\in \mathbb{C}:|z|<1\}$. The open unit disc of the complex plane is denoted by $\mathcal{H}\left(a,n\right)$. The subclass of $\mathcal{H}\left(U\right)$ of functions is denoted with the form $f\left(z\right)=a+a_n{z}^n+a_{n+1}{z}^{n+1}+\cdots$ and ${\mathcal{A}}_n=\{f\in \mathcal{H}\left(U\right):f\left(z\right)=z+a_{n+1}{z}^{n+1}+\cdots ,z\in U\},$ where $\mathcal{A}={\mathcal{A}}_1$.

The special class of starlike functions of order $\alpha$ is defined as

$${\mathcal{S}}^{*}\left(\alpha \right)=\left\{f\in \mathcal{A}:\mathrm{Re}\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}>\alpha ,0\le \alpha <1\right\}$$

and the class of convex functions of order $\alpha$ is defined as

$$\mathcal{K}\left(\alpha \right)=\left\{f\in \mathcal{A}:\mathrm{Re}\left(\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}+1\right)>\alpha ,0\le \alpha <1\right\}.$$

For introducing the new operator, the following previously known results are necessary.

Definition 1

([21]). For $f\in \mathcal{A}$, $m\in \mathbb{N}\cup \left\{0\right\}$, $\alpha ,l\ge 0$, the multiplier transformation $I\left(m,\alpha ,l\right)f\left(z\right)$ is defined by the following infinite series:

$$I\left(m,\alpha ,l\right)f\left(z\right):=z+\sum _{k=2}^{\infty }{\left(\frac{1+\alpha \left(k-1\right)+l}{1+l}\right)}^m{a}_k{z}^k.$$

We recall that the Riemann–Liouville fractional integral ([22]) is defined by the following relationship:

$${}_c^{RL}{I}_z^{\nu }f\left(z\right)=\frac{1}{\mathrm{\Gamma }\left(\nu \right)}{\int }_c^z{\left(z-w\right)}^{\nu -1}f\left(w\right)dw,\mathrm{Re}\left(\nu \right)>0,$$

which is used in the Atangana–Baleanu fractional integral.

In paper [23], symmetric and anti-symmetric derivatives of the Riemann–Liouville and Caputo type were defined, and the reflection symmetry properties of fractional differentiation were studied.

Definition 2

([24]). Let c be a fixed complex number and f be a complex function which is analytic on an open star-domain D centered at c. The extended Atangana–Baleanu integral, denoted by ${}_c^{AB}{I}_z^{\nu }f\left(z\right)$, is defined for any $\nu \in \mathbb{C}$ and any $z\in D\setminus \left\{c\right\}$ by:

$${}_c^{AB}{I}_z^{\nu }f\left(z\right)=\frac{1-\nu }{B\left(\nu \right)}f\left(z\right)+\frac{\nu }{B\left(\nu \right)}{}_c^{RL}{I}_z^{\nu }f\left(z\right).$$

Proposition 1

([24]). The extended Atangana–Baleanu integral from Definition 2 is an analytic function of both $z\in D\setminus \left\{c\right\}$ and $\nu \in \mathbb{C}$, provided f and B are analytic and B is nonzero. This is identical to the original formula in the real case when $0<\nu <1$ and $c<z$ in $\mathbb{R}$.

Therefore, it provides the analytic continuation of the original Atangana–Baleanu integral to complex values of z and ν.

Applying the Atangana–Baleanu fractional integral for $c=0$ to multiplier transformation, we define a new operator:

Definition 3.

Let $f\in \mathcal{A}$, $m\in \mathbb{N}\cup \left\{0\right\}$, $\alpha ,l\ge 0$, $\nu \in \mathbb{C}$, and any $z\in D\setminus \left\{0\right\}.$ The Atangana–Baleanu fractional integral associated with the multiplier transformation $I\left(m,\alpha ,l\right)f$ is defined by:

$${}_0^{AB}{I}_z^{\nu }\left(I\left(m,\alpha ,l\right)f\left(z\right)\right)=\frac{1-\nu }{B\left(\nu \right)}I\left(m,\alpha ,l\right)f\left(z\right)+\frac{\nu }{B\left(\nu \right)}{}_0^{RL}{I}_z^{\nu }I\left(m,\alpha ,l\right)f\left(z\right).$$

After a simple calculation, the following form is obtained for this operator:

$${}_0^{AB}{I}_z^{\nu }\left(I\left(m,\alpha ,l\right)f\left(z\right)\right)=\frac{1-\nu }{B\left(\nu \right)}z+\frac{\nu }{B\left(\nu \right)\mathrm{\Gamma }\left(\nu +2\right)}{z}^{\nu +1}+$$

$$\frac{1-\nu }{B\left(\nu \right)}\sum _{k=2}^{\infty }{\left(\frac{1+\alpha \left(k-1\right)+l}{l+1}\right)}^m{a}_k{z}^k+\frac{\nu }{B\left(\nu \right)}\sum _{k=2}^{\infty }{\left(\frac{1+\alpha \left(k-1\right)+l}{l+1}\right)}^m\frac{\mathrm{\Gamma }\left(k+1\right)}{\nu +k+1}{a}_k{z}^{k+\nu },$$

for the function $f\left(z\right)=z+{\sum }_{k=2}^{\infty }{a}_k{z}^k\in \mathcal{A}$.

The new class is defined using the new operator.

Definition 4.

A function $f\in \mathcal{A}$ is said to be in the class ${}_0^{AB}\mathcal{I}(m,\alpha ,l,\nu ,\gamma ,d,\beta )$ if it satisfies the following criterion:

$$\left|\frac{1}{d}\left(\frac{z{{(}_0^{AB}{I}_z^{\nu }\left(I\left(m,\alpha ,l\right)f\left(z\right)\right))}^{\prime }+\gamma z^2{{(}_0^{AB}{I}_z^{\nu }\left(I\left(m,\alpha ,l\right)f\left(z\right)\right))}^{″}}{{(1-\gamma )}_0^{AB}{I}_z^{\nu }\left(I\left(m,\alpha ,l\right)f\left(z\right)\right)+\gamma z{{(}_0^{AB}{I}_z^{\nu }\left(I\left(m,\alpha ,l\right)f\left(z\right)\right))}^{\prime }}-1\right)\right|<\beta ,$$

where $m\in \mathbb{N}\cup \left\{0\right\}$, $\alpha ,l\ge 0$, $\nu \in \mathbb{C},d$$\in \mathbb{C}-\left\{0\right\},$$0<\beta \le 1,$$0\le \gamma \le 1,$$z\in U\setminus \left\{0\right\}$.

In this section, a new subclass of analytic functions was introduced in Definition 4 after we presented the notations and definitions used during our investigation. The properties regarding the coefficient inequalities for the functions contained in the newly introduced class are obtained in Section 2 of the paper. Distortion bounds for functions from the class and for their derivative are given in Section 3, and properties regarding closure of the class are proven in Section 4, considering partial sums of functions from the class, with extreme points of the class being also provided. In Section 5, inclusion relations are obtained for certain values of the parameters involved, and neighborhood properties are discussed, while radii of starlikeness, convexity and close-to-convexity of the class are obtained in Section 6 of the paper.

To shorten the formulas, we have to make the notation $A={\left(\frac{1+\alpha \left(k-1\right)+l}{l+1}\right)}^m$ throughout the paper.

2. Coefficient Inequality

Theorem 1.

The function $f\in \mathcal{A}$ belongs to the class ${}_0^{AB}\mathcal{I}(m,\alpha ,l,\nu ,\gamma ,d,\beta )$ if and only if

$$\sum _{k=2}^{\infty }A\left[\left(1-\nu \right)\left[\gamma \left(k-1\right)+1\right]\left(k-1+\beta \left|d\right|\right)+\right.$$

$$\left.\frac{\nu \left[\gamma \left(k+\nu -1\right)+1\right]\left(k+\nu -1+\beta \left|d\right|\right)\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\right]a_k\le$$

$$\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}.$$

(1)

where $m\in \mathbb{N}\cup \left\{0\right\}$, $\alpha ,l\ge 0$, $\nu \in \mathbb{C},d$$\in \mathbb{C}-\left\{0\right\},$$0<\beta \le 1,$$0\le \gamma \le 1,$$z\in U\setminus \left\{0\right\}$.

Proof. 

Let $f\in \mathcal{A}$. Consider the inequality (1). After making an easy calculus, we get:

$$\left|\frac{z{{(}_0^{AB}{I}_z^{\nu }\left(I\left(m,\alpha ,l\right)f\left(z\right)\right))}^{\prime }+\gamma z^2{{(}_0^{AB}{I}_z^{\nu }\left(I\left(m,\alpha ,l\right)f\left(z\right)\right))}^{″}}{{(1-\gamma )}_0^{AB}{I}_z^{\nu }\left(I\left(m,\alpha ,l\right)f\left(z\right)\right)+\gamma z{{(}_0^{AB}{I}_z^{\nu }\left(I\left(m,\alpha ,l\right)f\left(z\right)\right))}^{\prime }}-1\right|=$$

$$\left|\frac{\frac{{\nu }^2\left(1+\gamma \nu \right)}{B\left(\nu \right)\mathrm{\Gamma }\left(\nu +2\right)}{z}^{\nu +1}+\frac{1-\nu }{B\left(\nu \right)}{\sum }_{k=2}^{\infty }A\left(k-1\right)\left[\gamma \left(k-1\right)+1\right]a_k{z}^k+\frac{\nu }{B\left(\nu \right)}{\sum }_{k=2}^{\infty }A\frac{\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\left(k+\nu -1\right)\left[\gamma \left(k+\nu -1\right)+1\right]a_k{z}^{k+\nu }}{\frac{1-\nu }{B\left(\nu \right)}z+\frac{\nu \left(\gamma \nu +1\right)}{B\left(\nu \right)\mathrm{\Gamma }\left(\nu +2\right)}{z}^{\nu +1}+\frac{1-\nu }{B\left(\nu \right)}{\sum }_{k=2}^{\infty }A\left[\gamma \left(k-1\right)+1\right]a_k{z}^k+\frac{\nu }{B\left(\nu \right)}{\sum }_{k=2}^{\infty }A\frac{\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\left[\gamma \left(k+\nu -1\right)+1\right]a_k{z}^{k+\nu }}\right|\le$$

and applying properties of modulus function, we obtain:

$$\begin{array}{l}\frac{\frac{{\nu }^2\left(1+\gamma \nu \right)}{B\left(\nu \right)\mathrm{\Gamma }\left(\nu +2\right)}{\left|z\right|}^{\nu }+\frac{1-\nu }{B\left(\nu \right)}{\sum }_{k=2}^{\infty }A\left(k-1\right)\left[\gamma \left(k-1\right)+1\right]a_k{\left|z\right|}^{k-1}+\frac{\nu }{B\left(\nu \right)}{\sum }_{k=2}^{\infty }A\frac{\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\left(k+\nu -1\right)\left[\gamma \left(k+\nu -1\right)+1\right]a_k{\left|z\right|}^{k+\nu -1}}{\frac{1-\nu }{B\left(\nu \right)}-\frac{\nu \left(\gamma \nu +1\right)}{B\left(\nu \right)\mathrm{\Gamma }\left(\nu +2\right)}{\left|z\right|}^{\nu }-\frac{1-\nu }{B\left(\nu \right)}{\sum }_{k=2}^{\infty }A\left[\gamma \left(k-1\right)+1\right]a_k{\left|z\right|}^{k-1}-\frac{\nu }{B\left(\nu \right)}{\sum }_{k=2}^{\infty }A\frac{\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\left[\gamma \left(k+\nu -1\right)+1\right]a_k{\left|z\right|}^{k+\nu -1}}\\ \le \beta \left|d\right|.\end{array}$$

Taking values of z on a real axis and for $z\to 1^{-}$, we find:

$$\sum _{k=2}^{\infty }A\left[\left(1-\nu \right)\left[\gamma \left(k-1\right)+1\right]\left(k-1+\beta \left|d\right|\right)+\right.$$

$$\left.\frac{\nu \left[\gamma \left(k+\nu -1\right)+1\right]\left(k+\nu -1+\beta \left|d\right|\right)\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\right]a_k\le$$

$$\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}.$$

Conversely, consider $f{\in }_0^{AB}\mathcal{I}(m,\alpha ,l,\nu ,\gamma ,d,\beta )$. Then, we obtain the inequality

$$Re\left\{\frac{z{{(}_0^{AB}{I}_z^{\nu }\left(I\left(m,\alpha ,l\right)f\left(z\right)\right))}^{\prime }+\gamma z^2{{(}_0^{AB}{I}_z^{\nu }\left(I\left(m,\alpha ,l\right)f\left(z\right)\right))}^{″}}{{(1-\gamma )}_0^{AB}{I}_z^{\nu }\left(I\left(m,\alpha ,l\right)f\left(z\right)\right)+\gamma z{{(}_0^{AB}{I}_z^{\nu }\left(I\left(m,\alpha ,l\right)f\left(z\right)\right))}^{\prime }}-1\right\}>-\beta \left|d\right|,$$

written as

$$\begin{array}{l}Re\left\{\frac{\frac{{\nu }^2\left(1+\gamma \nu \right)}{B\left(\nu \right)\mathrm{\Gamma }\left(\nu +2\right)}{z}^{\nu +1}+\frac{1-\nu }{B\left(\nu \right)}{\sum }_{k=2}^{\infty }A\left(k-1\right)\left[\gamma \left(k-1\right)+1\right]a_k{z}^k+\frac{\nu }{B\left(\nu \right)}{\sum }_{k=2}^{\infty }A\frac{\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\left(k+\nu -1\right)\left[\gamma \left(k+\nu -1\right)+1\right]a_k{z}^{k+\nu }}{\frac{1-\nu }{B\left(\nu \right)}z+\frac{\nu \left(\gamma \nu +1\right)}{B\left(\nu \right)\mathrm{\Gamma }\left(\nu +2\right)}{z}^{\nu +1}+\frac{1-\nu }{B\left(\nu \right)}{\sum }_{k=2}^{\infty }A\left[\gamma \left(k-1\right)+1\right]a_k{z}^k+\frac{\nu }{B\left(\nu \right)}{\sum }_{k=2}^{\infty }A\frac{\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\left[\gamma \left(k+\nu -1\right)+1\right]a_k{z}^{k+\nu }}\right.\\ \left.+\beta \left|d\right|\right\}>0\end{array}$$

and equivalently with:

$$\begin{array}{c}Re\left\{\frac{\frac{\left(1-\nu \right)\beta \left|d\right|}{B\left(\nu \right)}z+\frac{\nu \left(1+\gamma \nu \right)\left(\nu +\beta \left|d\right|\right)}{B\left(\nu \right)\mathrm{\Gamma }\left(\nu +2\right)}{z}^{\nu +1}+\frac{1-\nu }{B\left(\nu \right)}{\sum }_{k=2}^{\infty }A\left(k-1+\beta \left|d\right|\right)\left[\gamma \left(k-1\right)+1\right]a_k{z}^k}{\frac{1-\nu }{B\left(\nu \right)}z+\frac{\nu \left(\gamma \nu +1\right)}{B\left(\nu \right)\mathrm{\Gamma }\left(\nu +2\right)}{z}^{\nu +1}+\frac{1-\nu }{B\left(\nu \right)}{\sum }_{k=2}^{\infty }A\left[\gamma \left(k-1\right)+1\right]a_k{z}^k+\frac{\nu }{B\left(\nu \right)}{\sum }_{k=2}^{\infty }A\frac{\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\left[\gamma \left(k+\nu -1\right)+1\right]a_k{z}^{k+\nu }}+\right.\\ \left.\frac{\frac{\nu }{B\left(\nu \right)}{\sum }_{k=2}^{\infty }A\frac{\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\left(k+\nu -1+\beta \left|d\right|\right)\left[\gamma \left(k+\nu -1\right)+1\right]a_k{z}^{k+\nu }}{\frac{1-\nu }{B\left(\nu \right)}z+\frac{\nu \left(\gamma \nu +1\right)}{B\left(\nu \right)\mathrm{\Gamma }\left(\nu +2\right)}{z}^{\nu +1}+\frac{1-\nu }{B\left(\nu \right)}{\sum }_{k=2}^{\infty }A\left[\gamma \left(k-1\right)+1\right]a_k{z}^k+\frac{\nu }{B\left(\nu \right)}{\sum }_{k=2}^{\infty }A\frac{\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\left[\gamma \left(k+\nu -1\right)+1\right]a_k{z}^{k+\nu }}\right\}>0.\end{array}$$

Considering that $Re(-e^{i\theta })\ge -\left|e^{i\theta }\right|=-1$, the inequality becomes:

$$\begin{array}{l}\frac{\frac{\left(1-\nu \right)\beta \left|d\right|}{B\left(\nu \right)}-\frac{\nu \left(1+\gamma \nu \right)\left(\nu +\beta \left|d\right|\right)}{B\left(\nu \right)\mathrm{\Gamma }\left(\nu +2\right)}{r}^{\nu }-\frac{1-\nu }{B\left(\nu \right)}{\sum }_{k=2}^{\infty }A\left(k-1+\beta \left|d\right|\right)\left[\gamma \left(k-1\right)+1\right]a_k{r}^{k-1}}{\frac{1-\nu }{B\left(\nu \right)}-\frac{\nu \left(\gamma \nu +1\right)}{B\left(\nu \right)\mathrm{\Gamma }\left(\nu +2\right)}{r}^{\nu }-\frac{1-\nu }{B\left(\nu \right)}{\sum }_{k=2}^{\infty }A\left[\gamma \left(k-1\right)+1\right]a_k{r}^{k-1}-\frac{\nu }{B\left(\nu \right)}{\sum }_{k=2}^{\infty }A\frac{\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\left[\gamma \left(k+\nu -1\right)+1\right]a_k{r}^{k+\nu -1}}-\\ \frac{\frac{\nu }{B\left(\nu \right)}{\sum }_{k=2}^{\infty }A\frac{\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\left(k+\nu -1+\beta \left|d\right|\right)\left[\gamma \left(k+\nu -1\right)+1\right]a_k{r}^{k+\nu -1}}{\frac{1-\nu }{B\left(\nu \right)}-\frac{\nu \left(\gamma \nu +1\right)}{B\left(\nu \right)\mathrm{\Gamma }\left(\nu +2\right)}{r}^{\nu }-\frac{1-\nu }{B\left(\nu \right)}{\sum }_{k=2}^{\infty }A\left[\gamma \left(k-1\right)+1\right]a_k{r}^{k-1}-\frac{\nu }{B\left(\nu \right)}{\sum }_{k=2}^{\infty }A\frac{\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\left[\gamma \left(k+\nu -1\right)+1\right]a_k{r}^{k+\nu -1}}>0.\end{array}$$

Applying the mean value theorem when $r\to 1^{-}$ we get the inequality (1), and the proof is complete. □

Corollary 1.

Coefficients of the function $f{\in }_0^{AB}\mathcal{I}(m,\alpha ,l,\nu ,\gamma ,d,\beta )$ have the property:

$$a_k\le \frac{\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}}{A\left[\left(1-\nu \right)\left[\gamma \left(k-1\right)+1\right]\left(k-1+\beta \left|d\right|\right)+\frac{\nu \left[\gamma \left(k+\nu -1\right)+1\right]\left(k+\nu -1+\beta \left|d\right|\right)\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\right]},$$

for $k\ge 2.$

3. Distortion Theorems

Theorem 2.

Thefunction $f{\in }_0^{AB}\mathcal{I}(m,\alpha ,l,\nu ,\gamma ,d,\beta )$, for $\left|z\right|=r<1$, has the property:

$$r-\frac{\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}}{{\left(\frac{1+\alpha +l}{l+1}\right)}^m\left[\left(1-\nu \right)\left(\gamma +1\right)\left(\beta \left|d\right|+1\right)+\frac{2\nu \left[\gamma \left(\nu +1\right)+1\right]\left(\nu +1+\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +3\right)}\right]}{r}^2\le \left|f\left(z\right)\right|$$

$$\le r+\frac{\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}}{{\left(\frac{1+\alpha +l}{l+1}\right)}^m\left[\left(1-\nu \right)\left(\gamma +1\right)\left(\beta \left|d\right|+1\right)+\frac{2\nu \left[\gamma \left(\nu +1\right)+1\right]\left(\nu +1+\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +3\right)}\right]}{r}^2.$$

The equality holds for the function

$$f\left(z\right)=z+\frac{\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}}{{\left(\frac{1+\alpha +l}{l+1}\right)}^m\left[\left(1-\nu \right)\left(\gamma +1\right)\left(\beta \left|d\right|+1\right)+\frac{2\nu \left[\gamma \left(\nu +1\right)+1\right]\left(\nu +1+\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +3\right)}\right]}{z}^2,z\in U.$$

(2)

Proof. 

Let $f{\in }_0^{AB}\mathcal{I}(m,\alpha ,l,\nu ,\gamma ,d,\beta )$. Then, considering the relation (1) and that the sequence

$$\sum _{k=2}^{\infty }A\left[\left(1-\nu \right)\left[\gamma \left(k-1\right)+1\right]\left(k-1+\beta \left|d\right|\right)+\frac{\nu \left[\gamma \left(k+\nu -1\right)+1\right]\left(k+\nu -1+\beta \left|d\right|\right)\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\right]$$

is increasing and positive for $k\ge 2$, we obtain the inequality

$${\left(\frac{1+\alpha +l}{l+1}\right)}^m\left[\left(1-\nu \right)\left(\gamma +1\right)\left(\beta \left|d\right|+1\right)+\frac{2\nu \left[\gamma \left(\nu +1\right)+1\right]\left(\nu +1+\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +3\right)}\right]\sum _{k=2}^{\infty }{a}_k\le$$

$$\sum _{k=2}^{\infty }A\left[\left(1-\nu \right)\left[\gamma \left(k-1\right)+1\right]\left(k-1+\beta \left|d\right|\right)+\right.$$

$$\left.\frac{\nu \left[\gamma \left(k+\nu -1\right)+1\right]\left(k+\nu -1+\beta \left|d\right|\right)\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\right]a_k\le$$

$$\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)},$$

equivalently with

$$\sum _{k=2}^{\infty }{a}_k\le \frac{\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}}{{\left(\frac{1+\alpha +l}{l+1}\right)}^m\left[\left(1-\nu \right)\left(\gamma +1\right)\left(\beta \left|d\right|+1\right)+\frac{2\nu \left[\gamma \left(\nu +1\right)+1\right]\left(\nu +1+\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +3\right)}\right]}.$$

(3)

Using the properties of modulus function for

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

we get

$$r-r^2\sum _{k=2}^{\infty }{a}_k\le r-\sum _{k=2}^{\infty }{a}_k{r}^k\le \left|z\right|-\sum _{k=2}^{\infty }{a}_k{\left|z\right|}^k\le \left|f\left(z\right)\right|$$

$$\le \left|z\right|+\sum _{k=2}^{\infty }{a}_k{\left|z\right|}^k\le r+\sum _{k=2}^{\infty }{a}_k{r}^k\le r+r^2\sum _{k=2}^{\infty }{a}_k,$$

and considering relation (3), we obtain

$$r-\frac{\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}}{{\left(\frac{1+\alpha +l}{l+1}\right)}^m\left[\left(1-\nu \right)\left(\gamma +1\right)\left(\beta \left|d\right|+1\right)+\frac{2\nu \left[\gamma \left(\nu +1\right)+1\right]\left(\nu +1+\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +3\right)}\right]}{r}^2\le \left|f\left(z\right)\right|$$

$$\le r+\frac{\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}}{{\left(\frac{1+\alpha +l}{l+1}\right)}^m\left[\left(1-\nu \right)\left(\gamma +1\right)\left(\beta \left|d\right|+1\right)+\frac{2\nu \left[\gamma \left(\nu +1\right)+1\right]\left(\nu +1+\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +3\right)}\right]}{r}^2,$$

completing the proof. □

Theorem 3.

The function $f\in$${}_0^{AB}\mathcal{I}(m,\alpha ,l,\nu ,\gamma ,d,\beta )$, for $\left|z\right|=r<1$, has the property:

$$1-\frac{2\left[\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}\right]}{{\left(\frac{1+\alpha +l}{l+1}\right)}^m\left[\left(1-\nu \right)\left(\gamma +1\right)\left(\beta \left|d\right|+1\right)+\frac{2\nu \left[\gamma \left(\nu +1\right)+1\right]\left(\nu +1+\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +3\right)}\right]}r\le \left|f^{\prime }\left(z\right)\right|$$

$$\le 1+\frac{2\left[\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}\right]}{{\left(\frac{1+\alpha +l}{l+1}\right)}^m\left[\left(1-\nu \right)\left(\gamma +1\right)\left(\beta \left|d\right|+1\right)+\frac{2\nu \left[\gamma \left(\nu +1\right)+1\right]\left(\nu +1+\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +3\right)}\right]}r.$$

The equality holds for the function given by relation (2).

Proof. 

Using the properties of the modulus function for

$$f^{\prime }\left(z\right)=1+\sum _{k=2}^{\infty }k{a}_k{z}^{k-1},$$

we obtain:

$$1-\sum _{k=2}^{\infty }k{a}_k\left|z\right|\le 1-\sum _{k=2}^{\infty }k{a}_k{\left|z\right|}^{k-1}\le |f^{\prime }\left(z\right)|\le 1+\sum _{k=2}^{\infty }k{a}_k{\left|z\right|}^{k-1}\le 1+\sum _{k=2}^{\infty }k{a}_k\left|z\right|.$$

Applying relation (3), we get

$$1-\frac{2\left[\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}\right]}{{\left(\frac{1+\alpha +l}{l+1}\right)}^m\left[\left(1-\nu \right)\left(\gamma +1\right)\left(\beta \left|d\right|+1\right)+\frac{2\nu \left[\gamma \left(\nu +1\right)+1\right]\left(\nu +1+\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +3\right)}\right]}r\le \left|f^{\prime }\left(z\right)\right|$$

$$\le 1+\frac{2\left[\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}\right]}{{\left(\frac{1+\alpha +l}{l+1}\right)}^m\left[\left(1-\nu \right)\left(\gamma +1\right)\left(\beta \left|d\right|+1\right)+\frac{2\nu \left[\gamma \left(\nu +1\right)+1\right]\left(\nu +1+\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +3\right)}\right]}r,$$

and the proof is complete. □

4. Closure Theorems

Theorem 4.

The functions $f_p{\in }_0^{AB}\mathcal{I}(m,\alpha ,l,\nu ,\gamma ,d,\beta )$ of the following form

$$f_p\left(z\right)=z+\sum _{k=2}^{\infty }{a}_{k,p}{z}^k,a_{k,p}\ge 0,z\in U,$$

(4)

$p=1,2,\dots ,q,$ define the function h by relation

$$h\left(z\right)=\sum _{p=1}^q{\mu }_p{f}_p\left(z\right),{\mu }_p\ge 0,z\in U,$$

which belongs to the class ${}_0^{AB}\mathcal{I}(m,\alpha ,l,\nu ,\gamma ,d,\beta )$, when

$$\sum _{p=1}^q{\mu }_p=1.$$

Proof. 

The function h has the following form:

$$h\left(z\right)=\sum _{p=1}^q{\mu }_{p}z+\sum _{p=1}^q\sum _{k=2}^{\infty }{\mu }_p{a}_{k,p}{z}^k=z+\sum _{k=2}^{\infty }\sum _{p=1}^q{\mu }_p{a}_{k,p}{z}^k.$$

Regarding the functions $f_p$, $p=1,2,\dots ,q,$ being contained in the class ${}_0^{AB}\mathcal{I}(m,\alpha ,l,\nu ,\gamma ,d,\beta )$, by applying Theorem 1 we obtain:

$$\sum _{k=2}^{\infty }A\left[\left(1-\nu \right)\left[\gamma \left(k-1\right)+1\right]\left(k-1+\beta \left|d\right|\right)+\right.$$

$$\left.\frac{\nu \left[\gamma \left(k+\nu -1\right)+1\right]\left(k+\nu -1+\beta \left|d\right|\right)\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\right]a_{k,p}\le$$

$$\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}.$$

In this condition, we have to prove that:

$$\sum _{k=2}^{\infty }A\left[\left(1-\nu \right)\left[\gamma \left(k-1\right)+1\right]\left(k-1+\beta \left|d\right|\right)+\right.$$

$$\left.\frac{\nu \left[\gamma \left(k+\nu -1\right)+1\right]\left(k+\nu -1+\beta \left|d\right|\right)\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\right]\left(\sum _{p=1}^q{\mu }_p{a}_{k,p}\right)\le$$

$$\sum _{p=1}^q{\mu }_p\sum _{k=2}^{\infty }A\left[\left(1-\nu \right)\left[\gamma \left(k-1\right)+1\right]\left(k-1+\beta \left|d\right|\right)+\right.$$

$$\left.\frac{\nu \left[\gamma \left(k+\nu -1\right)+1\right]\left(k+\nu -1+\beta \left|d\right|\right)\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\right]a_{k,p}\le$$

$$\sum _{p=1}^q{\mu }_p\left(\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}\right)=$$

$$\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}.$$

Hence the proof is complete. □

Corollary 2.

Let the functions $f_p,$$p=1,2,$ as defined by relation (4) from the class ${}_0^{AB}\mathcal{I}(m,\alpha ,l,\nu ,\gamma ,d,\beta )$. Then, the function h given by

$$h\left(z\right)=(1-\xi )f_1\left(z\right)+\xi f_2\left(z\right),0\le \xi \le 1,z\in U,$$

belongs to the class ${}_0^{AB}\mathcal{I}(m,\alpha ,l,\nu ,\gamma ,d,\beta )$, as well.

Theorem 5.

Let

$$f_k\left(z\right)=z+$$

$$\frac{\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}}{A\left[\left(1-\nu \right)\left[\gamma \left(k-1\right)+1\right]\left(k-1+\beta \left|d\right|\right)+\frac{\nu \left[\gamma \left(k+\nu -1\right)+1\right]\left(k+\nu -1+\beta \left|d\right|\right)\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\right]}{z}^k,$$

$k\ge 2,$$z\in U$, and

$$f_1\left(z\right)=z.$$

The function f belongs to the class ${}_0^{AB}\mathcal{I}(m,\alpha ,l,\nu ,\gamma ,d,\beta )$ if and only if it can be written as

$$f\left(z\right)=\sum _{k=1}^{\infty }{\mu }_k{f}_k\left(z\right),z\in U,$$

with ${\mu }_k\ge 0,$$k\ge 1$ and ${\sum }_{k=1}^{\infty }{\mu }_k=1.$

Proof. 

Consider the function

$$f\left(z\right)=\sum _{k=1}^{\infty }{\mu }_k{f}_k\left(z\right)=$$

$$z+\sum _{k=2}^{\infty }\frac{\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}}{A\left[\left(1-\nu \right)\left[\gamma \left(k-1\right)+1\right]\left(k-1+\beta \left|d\right|\right)+\frac{\nu \left[\gamma \left(k+\nu -1\right)+1\right]\left(k+\nu -1+\beta \left|d\right|\right)\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\right]}{\mu }_k{z}^k;$$

we get:

$$\sum _{k=2}^{\infty }\frac{A\left[\left(1-\nu \right)\left[\gamma \left(k-1\right)+1\right]\left(k-1+\beta \left|d\right|\right)+\frac{\nu \left[\gamma \left(k+\nu -1\right)+1\right]\left(k+\nu -1+\beta \left|d\right|\right)\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\right]}{\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}}·$$

$$\frac{\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}}{A\left[\left(1-\nu \right)\left[\gamma \left(k-1\right)+1\right]\left(k-1+\beta \left|d\right|\right)+\frac{\nu \left[\gamma \left(k+\nu -1\right)+1\right]\left(k+\nu -1+\beta \left|d\right|\right)\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\right]}{\mu }_k$$

$$=\sum _{k=2}^{\infty }{\mu }_k=1-{\mu }_1\le 1.$$

Therefore, $f{\in }_0^{AB}\mathcal{I}(m,\alpha ,l,\nu ,\gamma ,d,\beta )$.

Conversely, considering $f{\in }_0^{AB}\mathcal{I}(m,\alpha ,l,\nu ,\gamma ,d,\beta )$ and the setting

$${\mu }_k=\frac{A\left[\left(1-\nu \right)\left[\gamma \left(k-1\right)+1\right]\left(k-1+\beta \left|d\right|\right)+\frac{\nu \left[\gamma \left(k+\nu -1\right)+1\right]\left(k+\nu -1+\beta \left|d\right|\right)\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\right]}{\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}}{a}_k,$$

and

$${\mu }_1=1-\sum _{k=2}^{\infty }{\mu }_k,$$

we obtain

$$f\left(z\right)=\sum _{k=1}^{\infty }{\mu }_k{f}_k\left(z\right).$$

Hence, the proof is complete. □

Corollary 3.

The class ${}_0^{AB}\mathcal{I}(m,\alpha ,l,\nu ,\gamma ,d,\beta )$ has as the extreme points the functions

$$f_1\left(z\right)=z,$$

and

$$f_k\left(z\right)=z+$$

$$\frac{\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}}{A\left[\left(1-\nu \right)\left[\gamma \left(k-1\right)+1\right]\left(k-1+\beta \left|d\right|\right)+\frac{\nu \left[\gamma \left(k+\nu -1\right)+1\right]\left(k+\nu -1+\beta \left|d\right|\right)\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\right]}{z}^k,$$

$k\ge 2,$$z\in U.$

5. Inclusion and Neighborhood Results

The $\delta$- neighborhood for a function $f\in \mathcal{A}$ is given by

$$N_{\delta }\left(f\right)=\{g\in \mathcal{A}:g\left(z\right)=z+\sum _{k=2}^{\infty }{b}_k{z}^k\mathrm{and}\sum _{k=2}^{\infty }k|a_k-b_k|\le \delta \},$$

(5)

and for $e\left(z\right)=z$, we obtain

$$N_{\delta }\left(e\right)=\{g\in \mathcal{A}:g\left(z\right)=z+\sum _{k=2}^{\infty }{b}_k{z}^k\mathrm{and}\sum _{k=2}^{\infty }k|b_k|\le \delta \}.$$

(6)

A function $f\in \mathcal{A}$ belongs to the class ${}_0^{AB}{\mathcal{I}}^{\zeta }(m,\alpha ,l,\nu ,\gamma ,d,\beta )$ if there exists a function $h{\in }_0^{AB}\mathcal{I}(m,\alpha ,l,\nu ,\gamma ,d,\beta )$ such that

$$\left|\frac{f\left(z\right)}{h\left(z\right)}-1\right|<1-\zeta ,z\in U,0\le \zeta <1.$$

(7)

Theorem 6.

We have the inclusion

$${}_0^{AB}\mathcal{I}(m,\alpha ,l,\nu ,\gamma ,d,\beta )\subset N_{\delta }\left(e\right)$$

for

$$\delta =\frac{2\left[\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}\right]}{{\left(\frac{1+\alpha +l}{l+1}\right)}^m\left[\left(1-\nu \right)\left(\gamma +1\right)\left(\beta \left|d\right|+1\right)+\frac{2\nu \left[\gamma \left(\nu +1\right)+1\right]\left(\nu +1+\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +3\right)}\right]}.$$

Proof. 

Consider $f{\in }_0^{AB}\mathcal{I}(m,\alpha ,l,\nu ,\gamma ,d,\beta )$. Using Theorem 1 and using the fact that the sequence

$$\sum _{k=2}^{\infty }A\left[\left(1-\nu \right)\left[\gamma \left(k-1\right)+1\right]\left(k-1+\beta \left|d\right|\right)+\frac{\nu \left[\gamma \left(k+\nu -1\right)+1\right]\left(k+\nu -1+\beta \left|d\right|\right)\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\right]$$

is increasing and positive for $k\ge 2$, as we say in Theorem 2, we obtain

$${\left(\frac{1+\alpha +l}{l+1}\right)}^m\left[\left(1-\nu \right)\left(\gamma +1\right)\left(\beta \left|d\right|+1\right)+\frac{2\nu \left[\gamma \left(\nu +1\right)+1\right]\left(\nu +1+\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +3\right)}\right]\sum _{k=2}^{\infty }{a}_k\le$$

$$\sum _{k=2}^{\infty }A\left[\left(1-\nu \right)\left[\gamma \left(k-1\right)+1\right]\left(k-1+\beta \left|d\right|\right)+\right.$$

$$\left.\frac{\nu \left[\gamma \left(k+\nu -1\right)+1\right]\left(k+\nu -1+\beta \left|d\right|\right)\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\right]a_k\le$$

$$\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)},$$

which implies

$$\sum _{k=2}^{\infty }{a}_k\le \frac{\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}}{{\left(\frac{1+\alpha +l}{l+1}\right)}^m\left[\left(1-\nu \right)\left(\gamma +1\right)\left(\beta \left|d\right|+1\right)+\frac{2\nu \left[\gamma \left(\nu +1\right)+1\right]\left(\nu +1+\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +3\right)}\right]}.$$

(8)

Applying Theorem 1 in conjunction with (8), we get

$$\sum _{k=2}^{\infty }k{a}_k\le \frac{2\left[\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}\right]}{{\left(\frac{1+\alpha +l}{l+1}\right)}^m\left[\left(1-\nu \right)\left(\gamma +1\right)\left(\beta \left|d\right|+1\right)+\frac{2\nu \left[\gamma \left(\nu +1\right)+1\right]\left(\nu +1+\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +3\right)}\right]}=\delta .$$

By virtue of (5), we obtain $f\in N_{\delta }\left(e\right)$, which completes the proof. □

Theorem 7.

If $h{\in }_0^{AB}\mathcal{I}(m,\alpha ,l,\nu ,\gamma ,d,\beta )$ and

$$\zeta =1-\frac{\delta }{2\left(1-\frac{\frac{\left(\beta -1\right)\left(\lambda +\left(1-\lambda \right)\mu \right)}{\left(\beta +1\right)\nu }-\frac{\beta \gamma B\left(\nu \right)}{\left(\beta +1\right)\left(1-\nu \right)\nu }-\frac{\lambda +\left(1+\nu -\lambda \right)\mu }{\left(1-\nu \right)\mathrm{\Gamma }\left(\nu +2\right)}}{{\left(\frac{1+\alpha +l}{l+1}\right)}^m\left[\frac{\lambda +\left(2-\lambda \right)\mu }{\nu }+\frac{2\left(\lambda +\left(2+\nu -\lambda \right)\mu \right)}{\left(1-\nu \right)\mathrm{\Gamma }\left(\nu +3\right)}\right]}\right)},$$

(9)

then

$$N_{\delta }\left(h\right){\subset }_0^{AB}{\mathcal{I}}^{\zeta }(m,\alpha ,l,\nu ,\gamma ,d,\beta ).$$

Proof. 

Consider $f\in N_{\delta }\left(h\right)$. Relation (5)

$$\sum _{k=2}^{\infty }k|a_k-b_k|\le \delta$$

implies

$$\sum _{k=2}^{\infty }|a_k-b_k|\le \frac{\delta }{2}.$$

(10)

Using relation (8) for $h{\in }_0^{AB}\mathcal{I}(m,\alpha ,l,\nu ,\gamma ,d,\beta ),$ we get:

$$\sum _{k=2}^{\infty }{b}_k\le \frac{\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}}{{\left(\frac{1+\alpha +l}{l+1}\right)}^m\left[\left(1-\nu \right)\left(\gamma +1\right)\left(\beta \left|d\right|+1\right)+\frac{2\nu \left[\gamma \left(\nu +1\right)+1\right]\left(\nu +1+\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +3\right)}\right]}.$$

(11)

Using (10) and (11), we have:

$$\left|\frac{f\left(z\right)}{h\left(z\right)}-1\right|\le \frac{{\sum }_{k=2}^{\infty }|a_k-b_k|}{1-{\sum }_{k=2}^{\infty }{b}_k}\le$$

$$\frac{\delta }{2\left(1-\frac{\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}}{{\left(\frac{1+\alpha +l}{l+1}\right)}^m\left[\left(1-\nu \right)\left(\gamma +1\right)\left(\beta \left|d\right|+1\right)+\frac{2\nu \left[\gamma \left(\nu +1\right)+1\right]\left(\nu +1+\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +3\right)}\right]}\right)}=1-\zeta .$$

By relation (7), we obtain $f{\in }_0^{AB}{\mathcal{I}}^{\zeta }(m,\alpha ,l,\nu ,\gamma ,d,\beta )$, where $\zeta$ is given by (9). □

6. Radii of Starlikeness, Convexity and Close-to-Convexity

Theorem 8.

The function $f{\in }_0^{AB}\mathcal{I}(m,\alpha ,l,\nu ,\gamma ,d,\beta )$ is univalent starlike of order δ, $0\le \delta <1,$ in $\left|z\right|<r_1$, with

$r_1={inf}_k{\left\{\frac{\left(1-\delta \right)A\left[\left(1-\nu \right)\left[\gamma \left(k-1\right)+1\right]\left(k-1+\beta \left|d\right|\right)+\frac{\nu \left[\gamma \left(k+\nu -1\right)+1\right]\left(k+\nu -1+\beta \left|d\right|\right)\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\right]}{\left(k-\delta \right)\left[\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}\right]}\right\}}^{\frac{1}{k-1}}.$

For the function of the form

$$f_k\left(z\right)=z+$$

$$\frac{\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}}{A\left[\left(1-\nu \right)\left[\gamma \left(k-1\right)+1\right]\left(k-1+\beta \left|d\right|\right)+\frac{\nu \left[\gamma \left(k+\nu -1\right)+1\right]\left(k+\nu -1+\beta \left|d\right|\right)\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\right]}{z}^k,$$

(12)

$k\ge 2,$ the result is sharp.

Proof. 

For the function f to be univalent starlike of order $\delta$, we have to show that

$$\left|\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}-1\right|\le 1-\delta ,\left|z\right|<r_1.$$

For $f\in \mathcal{A}$, we can write

$$\left|\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}-1\right|=\left|\frac{{\sum }_{k=2}^{\infty }(k-1)a_k{z}^{k-1}}{1+{\sum }_{k=2}^{\infty }{a}_k{z}^{k-1}}|\right|\le \frac{{\sum }_{k=2}^{\infty }(k-1)a_k{\left|z\right|}^{k-1}}{1-{\sum }_{k=2}^{\infty }{a}_k{\left|z\right|}^{k-1}},$$

and it remains to be seen that

$$\frac{{\sum }_{k=2}^{\infty }(k-1)a_k{\left|z\right|}^{k-1}}{1-{\sum }_{k=2}^{\infty }{a}_k{\left|z\right|}^{k-1}}\le 1-\delta$$

equivalently to

$$\sum _{k=2}^{\infty }(k-\delta )a_k{\left|z\right|}^{k-1}\le 1-\delta .$$

Applying Theorem 1, we get

$${\left|z\right|}^{k-1}\le \frac{\left(1-\delta \right)A\left[\left(1-\nu \right)\left[\gamma \left(k-1\right)+1\right]\left(k-1+\beta \left|d\right|\right)+\frac{\nu \left[\gamma \left(k+\nu -1\right)+1\right]\left(k+\nu -1+\beta \left|d\right|\right)\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\right]}{\left(k-\delta \right)\left[\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}\right]},$$

or

$$\left|z\right|\le {\left\{\frac{\left(1-\delta \right)A\left[\left(1-\nu \right)\left[\gamma \left(k-1\right)+1\right]\left(k-1+\beta \left|d\right|\right)+\frac{\nu \left[\gamma \left(k+\nu -1\right)+1\right]\left(k+\nu -1+\beta \left|d\right|\right)\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\right]}{\left(k-\delta \right)\left[\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}\right]}\right\}}^{\frac{1}{k-1}}.$$

Hence, the proof is complete. □

Theorem 9.

The function $f{\in }_0^{AB}\mathcal{I}(m,\alpha ,l,\nu ,\gamma ,d,\beta )$ is univalent convex of order $\delta ,$$0\le \delta \le 1,$ in $\left|z\right|<r_2$, with

$r_2={inf}_k{\left\{\frac{\left(1-\delta \right)A\left[\left(1-\nu \right)\left[\gamma \left(k-1\right)+1\right]\left(k-1+\beta \left|d\right|\right)+\frac{\nu \left[\gamma \left(k+\nu -1\right)+1\right]\left(k+\nu -1+\beta \left|d\right|\right)\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\right]}{k(k-\delta )\left[\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}\right]}\right\}}^{\frac{1}{k-1}}.$

For the function of the form

$$f_k\left(z\right)=z+\frac{\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}}{A\left[\left(1-\nu \right)\left[\gamma \left(k-1\right)+1\right]\left(k-1+\beta \left|d\right|\right)+\frac{\nu \left[\gamma \left(k+\nu -1\right)+1\right]\left(k+\nu -1+\beta \left|d\right|\right)\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\right]}{z}^k,k\ge 2,$$

the result is sharp.

Proof. 

For the function f to be univalent convex of order $\delta$, we have to prove that

$$\left|\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right))}\right|\le 1-\delta ,\left|z\right|<r_2.$$

For $f\in \mathcal{A}$, we can write

$$\left|\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right|=\left|\frac{{\sum }_{k=2}^{\infty }k(k-1)a_k{z}^{k-1}}{1+{\sum }_{k=2}^{\infty }k{a}_k{z}^{k-1}}\right|\le \frac{{\sum }_{k=2}^{\infty }k(k-1)a_k{\left|z\right|}^{k-1}}{1-{\sum }_{k=2}^{\infty }k{a}_k{\left|z\right|}^{k-1}},$$

and it remains to be seen that

$$\frac{{\sum }_{k=2}^{\infty }k(k-1)a_k{\left|z\right|}^{k-1}}{1-{\sum }_{k=2}^{\infty }k{a}_k{\left|z\right|}^{k-1}}\le 1-\delta ,$$

$$\sum _{k=2}^{\infty }k(k-\delta )a_k{\left|z\right|}^{k-1}\le 1-\delta .$$

Using Theorem 1, we obtain

$${\left|z\right|}^{k-1}\le \frac{\left(1-\delta \right)A\left[\left(1-\nu \right)\left[\gamma \left(k-1\right)+1\right]\left(k-1+\beta \left|d\right|\right)+\frac{\nu \left[\gamma \left(k+\nu -1\right)+1\right]\left(k+\nu -1+\beta \left|d\right|\right)\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\right]}{k(k-\delta )\left[\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}\right]},$$

or

$$\left|z\right|\le {\left\{\frac{\left(1-\delta \right)A\left[\left(1-\nu \right)\left[\gamma \left(k-1\right)+1\right]\left(k-1+\beta \left|d\right|\right)+\frac{\nu \left[\gamma \left(k+\nu -1\right)+1\right]\left(k+\nu -1+\beta \left|d\right|\right)\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\right]}{k(k-\delta )\left[\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}\right]}\right\}}^{\frac{1}{k-1}},$$

and the proof is complete. □

Theorem 10.

The function $f{\in }_0^{AB}\mathcal{I}(m,\alpha ,l,\nu ,\gamma ,d,\beta )$ is univalent close-to-convex of order $\delta ,$$0\le \delta <1,$ in $\left|z\right|<r_3$, with

$$r_3=\underset{k}{inf}{\left\{\frac{\left(1-\delta \right)A\left[\left(1-\nu \right)\left[\gamma \left(k-1\right)+1\right]\left(k-1+\beta \left|d\right|\right)+\frac{\nu \left[\gamma \left(k+\nu -1\right)+1\right]\left(k+\nu -1+\beta \left|d\right|\right)\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\right]}{k\left[\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}\right]}\right\}}^{\frac{1}{k-1}}.$$

For the function of the form

$$f_k\left(z\right)=z+\frac{\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}}{A\left[\left(1-\nu \right)\left[\gamma \left(k-1\right)+1\right]\left(k-1+\beta \left|d\right|\right)+\frac{\nu \left[\gamma \left(k+\nu -1\right)+1\right]\left(k+\nu -1+\beta \left|d\right|\right)\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\right]}{z}^k,k\ge 2,$$

the result is sharp.

Proof. 

For the function f to be univalent close-to-convex of order $\delta$, we have to prove that

$$|f^{\prime }\left(z\right)-1|\le 1-\delta ,|z|<r_3.$$

For $f\in \mathcal{A}$, we can write

$$|f^{\prime }\left(z\right)-1|=\left|\sum _{k=2}^{\infty }k{a}_k{z}^{k-1}\right|\le \sum _{k=2}^{\infty }k{a}_k{\left|z\right|}^{k-1}\le 1-\delta$$

if ${\sum }_{k=2}^{\infty }\frac{k{a}_k}{1-\delta }{\left|z\right|}^{k-1}\le 1$. Using Theorem 1, the inequality holds true if

$${\left|z\right|}^{k-1}\le \frac{\left(1-\delta \right)A\left[\left(1-\nu \right)\left[\gamma \left(k-1\right)+1\right]\left(k-1+\beta \left|d\right|\right)+\frac{\nu \left[\gamma \left(k+\nu -1\right)+1\right]\left(k+\nu -1+\beta \left|d\right|\right)\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\right]}{k\left[\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}\right]}$$

or

$$\left|z\right|\le {\left\{\frac{\left(1-\delta \right)A\left[\left(1-\nu \right)\left[\gamma \left(k-1\right)+1\right]\left(k-1+\beta \left|d\right|\right)+\frac{\nu \left[\gamma \left(k+\nu -1\right)+1\right]\left(k+\nu -1+\beta \left|d\right|\right)\mathrm{\Gamma }\left(k+1\right)}{\mathrm{\Gamma }\left(\nu +k+1\right)}\right]}{k\left[\beta \left|d\right|\left(1-\gamma \right)-\frac{\nu \left(\gamma \nu +1\right)\left(\nu +\beta \left|d\right|\right)}{\mathrm{\Gamma }\left(\nu +2\right)}\right]}\right\}}^{\frac{1}{k-1}}.$$

Hence, the proof is complete. □

7. Conclusions

In this paper, a new operator ${}_0^{AB}{I}_z^{\nu }\left(I\left(m,\alpha ,l\right)f\right)$ is introduced in Definition 3, applying the Atangana–Baleanu fractional integral for $c=0$ to a multiplier transformation. A new class of analytic functions ${}_0^{AB}\mathcal{I}(m,\alpha ,l,\nu ,\gamma ,d,\beta )$ is defined in Definition 4 and follows a study of this class regarding coefficient inequality, distortion and closure theorems, inclusion and neighborhood results, radii of starlikeness, convexity and close-to-convexity.

The class introduced in this paper is interesting due to the operator used for introducing it, since this operator is part of the celebrated family of fractional integral operators, which have been much investigated in the recent years. In addition to symmetry properties, which could be investigated related to the newly defined operator, algebraic properties could also be added after further studies in this regard. Considering the starlikeness and convexity properties of the newly defined class, symmetry properties could be found for this class, having in mind the connection between convexity and symmetry. Additionally, subordination and superordination properties could be obtained by using the means of the theories of differential subordination and superordination on this class due to its univalence properties. The results obtained in this paper could be adapted in view of quantum calculus aspects as seen in [25,26].