# Inclusion relations for certain families of integral operators associated with conic regions

## Abstract

In this work, we introduce certain subclasses of analytic functions involving the integral operators that generalize the class of uniformly starlike, convex, and close-to-convex functions with respect to symmetric points. We then establish various inclusion relations for these newly defined classes.

## Introduction

Let $$\mathcal{A}$$ be the class of functions

$$f(z)=z+\sum_{n=2}^{\infty }a_{n}z^{n}$$
(1.1)

analytic in the open unit disc $$\mathfrak{A}=\{z\in \mathbb{C} : \vert z \vert <1 \}$$, and let $$\mathcal{S}$$ be the class of functions in $$\mathcal{A}$$ that are univalent in $$\mathfrak{A}$$. Also let $$\mathcal{S}^{\ast }$$, $$\mathcal{C}$$, $$\mathcal{K}$$, and $$\mathcal{C}^{\ast }$$ be the subclasses of $$\mathcal{A}$$ consisting of all functions that are starlike, convex, close-to-convex, and quasiconvex, respectively; for details, see .

Let f and g be analytic in $$\mathfrak{A}$$. We say that f is subordinate to g, written as $$f(z)\prec g(z)$$, if there exists a Schwarz function w that is analytic in $$\mathfrak{A}$$ with $$w(0)=0$$ and $$\vert w(z) \vert <1$$ ($$z\in \mathfrak{A}$$) and such that $$f(z)=g(w(z))$$. In particular, when g is univalent, then such a subordination is equivalent to $$f(0)=g(0)$$ and $$f(\mathfrak{A}) \subseteq g(\mathfrak{A})$$; see .

Two points A and $$A^{\prime }$$ are said to be symmetrical with respect to M if M is the midpoint of the line segment $$AA^{\prime }$$. Sakaguchi  introduced and studied the class $$\mathcal{S} _{s}^{\ast }$$ of starlike functions with respect to symmetrical points z and −z belonging to the open unit disc $$\mathfrak{A}$$. The class $$\mathcal{S}_{s}^{\ast }$$ includes the classes of convex and odd starlike functions with respect to the origin. It was shown  that a necessary and sufficient condition for $$f ( z ) \in \mathcal{S}_{s}^{\ast }$$ to be univalent and starlike with respect to symmetrical points in $$\mathfrak{A}$$ is that

$$\frac{2zf^{\prime }(z)}{f(z)-f(-z)}\in \mathcal{P}, \quad z\in \mathfrak{A}.$$

Das and Singh  defined the classes $$\mathcal{C}_{s}$$ of convex functions with respect to symmetrical points and showed that a necessary and sufficient condition for $$f ( z ) \in \mathcal{C}_{s}$$ is that

$$\frac{2 ( zf^{\prime }(z) ) ^{\prime }}{ ( f(z)-f(-z) ) ^{\prime }}\in \mathcal{P}, \quad z\in \mathfrak{A}.$$

It is also well known  that $$f ( z ) \in \mathcal{C}_{s}$$ if and only if $$zf ( z ) \in \mathcal{S} _{s}^{\ast }$$.

The classes $$k-\mathcal{CV}$$ and $$k-\mathcal{ST}$$ with $$k\geq 0$$ denote the famous classes of k-uniformly convex and k-starlike functions, respectively, introduced by Kanas and Wisniowska, respectively. For some details see [4,5,6,7].

Consider the domain

$$\varOmega _{k}= \bigl\{ u+iv;u>k\sqrt{ ( u-1 ) ^{2}+v^{2}} \bigr\} .$$
(1.2)

For fixed k, $$\varOmega _{k}$$ represents the conic region bounded successively by the imaginary axis ($$k=0$$), the right branch of a hyperbola ($$0< k<1$$), a parabola ($$k=1$$), and an ellipse ($$k>1$$). This domain was studied by Kanas [4,5,6]. The function $$p_{k}$$ with $$p_{k} ( 0 ) =1$$ and $$p_{k}^{\prime } ( 0 ) >0$$ plays the role of extremal and is given by

$$p_{k} ( z ) = \textstyle\begin{cases} \frac{1+z}{1-z}, &k=0, \\ 1+\frac{2}{\pi ^{2}} ( \log \frac{1+\sqrt{z}}{1-\sqrt{z}} ) ^{2}, &k=1, \\ 1+\frac{2}{1-k^{2}}\sinh ^{2} [ ( \frac{2}{\pi }\arccos k ) \operatorname{arc}\tanh \sqrt{z} ] ,&0< k< 1, \\ 1+\frac{1}{k^{2}-1}\sin [ \frac{\pi }{2R ( t ) } \int _{0}^{\frac{u ( z ) }{\sqrt{t}}}\frac{1}{\sqrt{1-x ^{2}}\sqrt{1- ( tx ) ^{2}}}\, dx ] +\frac{1}{k ^{2}-1},&k>1, \end{cases}$$
(1.3)

with $$u(z)=\frac{z-\sqrt{t}}{1-\sqrt{tz}}$$, $$t\in (0,1)$$, $$z\in E$$, and t chosen such that $$k=\cosh ( \frac{\pi R^{\prime }(t)}{4R(t)} )$$, where $$R(t)$$ is Legendre’s complete elliptic integral of the first kind, and $$R^{\prime }(t)$$ is the complementary integral of $$R(t)$$ (see [5, 6]). Let $$\mathcal{P}_{p_{k}}$$ denote the class of all functions $$p ( z )$$ that are analytic in E with $$p ( 0 ) =1$$ and $$p ( z ) \prec p_{k} ( z )$$ for $$z\in E$$. Clearly, we can see that $$\mathcal{P}_{p_{k}}\subset \mathcal{P}$$, where $$\mathcal{P}$$ is the class of functions with positive real parts (see ). More precisely,

$$\mathcal{P}_{p_{k}}\subset \mathcal{P} \biggl( \frac{k}{1+k} \biggr) \subset \mathcal{P}.$$

For more detail regarding conic domains and related classes, see [4,5,6, 8,9,10,11].

Recently, Noor  defined the classes $$k-\mathcal{ST}_{s}$$, $$k-\mathcal{UCV}_{s}$$, and $$k-\mathcal{UK}_{s}$$ of k-uniformly starlike, convex, and close to convex functions with respect to symmetrical points and studied various interesting properties for these classes.

We consider the following one-parameter families of integral operators:

\begin{aligned}& \mathcal{I}_{\beta }^{\alpha }f ( z ) =\frac{ ( \beta +1 ) ^{\alpha }}{\varGamma ( \alpha ) z^{\beta }} \int _{0}^{z}t^{\beta -1} \biggl( \log \frac{z}{t} \biggr) ^{\alpha -1}f ( t ) \,dt, \end{aligned}
(1.4)
\begin{aligned}& \mathfrak{L}_{\beta }^{\alpha }f ( z ) =\binom{\alpha + \beta }{\beta } \frac{\alpha }{z^{\beta }} \int _{0}^{z}t^{\beta -1} \biggl( 1- \frac{t}{z} \biggr) ^{\alpha -1}f ( t ) \,dt, \end{aligned}
(1.5)

and

$$\mathfrak{J}_{\beta }f ( z ) =\frac{\beta +1}{z^{\beta }} \int _{0}^{z}t^{\beta -1}f ( t ) \,dt,$$
(1.6)

where $$\alpha \geq 0$$, $$\beta >-1$$, and Γ is the familiar gamma function. We note that $$\mathfrak{J}_{\beta }:\mathcal{A}\rightarrow \mathcal{A}$$ defined by (1.6) is the generalized Bernardi operator introduced in  for $$\beta =1,2,3,\ldots$$ , and for any real number $$\beta >-1$$, this operator was studied by Owa and Srivastava [14, 15]. For the operators $$\mathfrak{L}_{\beta }^{\alpha }$$ and $$\mathcal{I}_{\beta } ^{\alpha }$$, we refer to [16, 17]. Also, for $$\alpha =1$$, we see that

$$\mathfrak{J}_{\beta }f ( z ) =\mathfrak{L}_{\beta }^{1}f ( z ) =\mathcal{I}_{\beta }^{1}f ( z ) .$$

We can represent these operators as follows:

\begin{aligned}& \begin{aligned}[b] \mathcal{I}_{\beta }^{\alpha }f ( z ) &=z+\sum _{n=2}^{ \infty } \biggl( \frac{\beta +1}{\beta +n} \biggr) ^{\alpha }a_{n}z ^{n} \\ &= \Biggl( z+\sum_{n=2}^{\infty } \biggl( \frac{\beta +1}{\beta +n} \biggr) ^{\alpha }z^{n} \Biggr) \ast f ( z ) , \end{aligned} \end{aligned}
(1.7)
\begin{aligned}& \begin{aligned}[b] \mathfrak{L}_{\beta }^{\alpha }f ( z ) &=z+\sum _{n=2}^{ \infty }\frac{\varGamma ( \beta +n ) \varGamma ( \alpha + \beta +1 ) }{\varGamma ( \alpha +\beta +n ) \varGamma ( \beta +1 ) }a_{n}z^{n} \\ &=\binom{\alpha +\beta }{\beta }z{}_{2}F_{1} ( 1,\beta ; \alpha +\beta ;z ) \ast f ( z ) , \end{aligned} \end{aligned}
(1.8)

and

$$\mathfrak{J}_{\beta }f ( z ) =z+\sum_{n=2}^{\infty } \biggl( \frac{\beta +1}{\beta +n} \biggr) a_{n}z^{n},$$
(1.9)

where $${}_{2}F_{1}$$ denotes the Gaussian hypergeometric function, and the symbol stands for the convolution (Hadamard product).

By (1.7) and (1.8) we can easily derive the identities

$$z \bigl( \mathcal{I}_{\beta }^{\alpha }f ( z ) \bigr) ^{\prime }= ( \beta +1 ) \mathcal{I}_{\beta }^{\alpha -1}f ( z ) -\beta \mathcal{I}_{\beta }^{\alpha }f ( z )$$
(1.10)

and

$$z \bigl( \mathfrak{L}_{\beta }^{\alpha }f ( z ) \bigr) ^{\prime }= ( \alpha +\beta ) \mathfrak{L}_{\beta }^{ \alpha -1}f ( z ) - ( \alpha + \beta -1 ) \mathfrak{L}_{\beta }^{\alpha }f ( z ) ,$$
(1.11)

where $$\alpha \geq 1$$ and $$\beta >-1$$. From (1.10) we have

$$\biggl[ \frac{1}{1+\beta }p ( z ) +\frac{\beta }{1+\beta } \biggr] = \frac{\mathcal{I}_{\beta }^{\alpha -1}f ( z ) }{\mathcal{I}_{\beta }^{\alpha }f ( z ) }$$

with

$$p ( z ) =\frac{z ( \mathcal{I}_{\beta }^{\alpha }f ( z ) ) ^{\prime }}{\mathcal{I}_{\beta }^{\alpha }f ( z ) }.$$

With the help of these integral operators, we now define the following classes.

### Definition 1.1

Let $$f ( z ) \in \mathcal{A}$$. Then $$f ( z ) \in k-\mathcal{ST}_{s} ( \alpha ,\beta )$$, $$\alpha \geq 0$$, $$\beta >-1$$, if $$\mathcal{I}_{\beta }^{\alpha }f ( z ) \in k-\mathcal{ST}_{s}$$ in $$\mathfrak{A}$$.

### Definition 1.2

Let $$f ( z ) \in \mathcal{A}$$. Then $$f ( z ) \in k-\mathcal{ST}_{s}^{\ast } ( \alpha ,\beta )$$, $$\alpha \geq 0$$, $$\beta >-1$$, if $$\mathfrak{L}_{\beta }^{\alpha }f ( z ) \in k-\mathcal{ST}_{s}$$ in $$\mathfrak{A}$$.

### Definition 1.3

Let $$f ( z ) \in \mathcal{A}$$. Then $$f ( z ) \in k-\mathcal{UK}_{s} ( \alpha ,\beta )$$, $$\alpha \geq 0$$, $$\beta >-1$$, if $$\mathcal{I}_{\beta }^{\alpha }f ( z ) \in k-\mathcal{UK}_{s}$$ in $$\mathfrak{A}$$.

### Definition 1.4

Let $$f ( z ) \in \mathcal{A}$$. Then $$f ( z ) \in k-\mathcal{UK}_{s}^{\ast } ( \alpha ,\beta )$$, $$\alpha \geq 0$$, $$\beta >-1$$, if $$\mathfrak{L}_{\beta }^{\alpha }f ( z ) \in k-\mathcal{UK}_{s}$$ in $$\mathfrak{A}$$.

## A set of lemmas

In this section, we give the following lemmas, which will be used in our investigation.

### Lemma 2.1

()

Let $$k\geq 0$$, and let $$\beta _{1},\gamma \in \mathbb{C}$$ be such that $$\beta _{1}\neq 0$$ and $$\mathfrak{Re} \{ \frac{\beta _{1}k}{k+1}+ \gamma \} >0$$. Suppose that $$p ( z )$$ is analytic in $$\mathfrak{A}$$ with $$p ( 0 ) =1$$ and satisfies

$$\biggl( p ( z ) +\frac{zp^{\prime } ( z ) }{ \beta _{1}p ( z ) +\gamma } \biggr) \prec p_{k} ( z )$$
(2.1)

and that $$q ( z )$$ is an analytic function satisfying

$$q ( z ) +\frac{zq^{\prime } ( z ) }{\beta _{1}q ( z ) +\gamma }=p_{k} ( z ) .$$
(2.2)

Then $$q ( z )$$ is univalent, $$p ( z ) \prec q ( z ) \prec p_{k} ( z )$$, and $$q ( z )$$ is the best dominant of (2.1) given as

$$q ( z ) = \biggl[ \beta _{1} \int _{0}^{1} \biggl( t^{ \beta _{1}+\gamma -1}\exp \int _{z}^{tz}\frac{p_{k} ( u ) -1}{u}\,du \biggr) \,dt \biggr] ^{-1}-\frac{\gamma }{\beta _{1}}.$$
(2.3)

### Lemma 2.2

()

Let $$\lambda ,\rho \in \mathbb{C}$$ be such that $$\lambda \neq 0$$, and let $$\phi (z)\in \mathcal{A}$$ be convex and univalent in $$\mathbb{U}$$ with $$\mathfrak{Re} \{ \lambda \phi (z)+\rho \} >0$$ ($$z\in \mathbb{U}$$). Also, let $$q(z)\in \mathcal{A}$$ and $$q(z)\prec \phi (z)$$. If $$p(z)$$ is analytic in $$\mathbb{U}$$ with $$p ( 0 ) =1$$ and satisfies

$$\biggl( p(z)+\frac{zp^{\prime }(z)}{\lambda q(z)+\rho } \biggr) \prec \phi (z),$$
(2.4)

then $$p(z)\prec \phi (z)$$.

## The main results and their consequences

Our first main result is stated as the following:

### Theorem 3.1

Let $$f ( z ) \in k-\mathcal{ST}_{s} ( \alpha ,\beta )$$. Then the odd function

$$\psi ( z ) =\frac{1}{2} \bigl[ f ( z ) -f ( -z ) \bigr] \in k- \mathcal{ST} ( \alpha ,\beta ) .$$

### Proof

Note that

$$\mathcal{I}_{\beta }^{\alpha }\psi ( z ) =\frac{1}{2} \bigl[ \mathcal{I}_{\beta }^{\alpha }f ( z ) -\mathcal{I} _{\beta }^{\alpha }f ( -z ) \bigr] .$$

We want to show that $$\mathcal{I}_{\beta }^{\alpha }\psi ( z ) \in k-\mathcal{ST}$$. Now, for $$f ( z ) \in k- \mathcal{ST}_{s} ( \alpha ,\beta )$$, this implies that $$\mathcal{I}_{\beta }^{\alpha }f ( z ) \in k-\mathcal{ST} _{s}$$. Then, for $$z\in \mathfrak{A}$$,

\begin{aligned} \frac{z ( \mathcal{I}_{\beta }^{\alpha }\psi ( z ) ) ^{{\prime }}}{\mathcal{I}_{\beta }^{\alpha }\psi ( z ) } =&\frac{1}{2} \biggl[ \frac{2z ( \mathcal{I}_{\beta }^{\alpha }f ( z ) ) ^{{\prime }}}{\mathcal{I}_{ \beta }^{\alpha }f ( z ) -\mathcal{I}_{\beta }^{\alpha }f ( -z ) }+ \frac{2 ( -z ) ( \mathcal{I} _{\beta }^{\alpha }f ( -z ) ) ^{{\prime }}}{ \mathcal{I}_{\beta }^{\alpha }f ( -z ) -\mathcal{I}_{ \beta }^{\alpha }f ( z ) } \biggr] \\ =&\frac{1}{2} \bigl[ h_{1} ( z ) +h_{2} ( z ) \bigr] \\ =&h ( z ) . \end{aligned}

and $$h_{i} ( z ) \prec p_{k} ( z )$$, $$i=1,2$$. This implies that $$h ( z ) \prec p_{k} ( z )$$ in $$\mathfrak{A}$$, and therefore $$\mathcal{I}_{\beta } ^{\alpha }\psi ( z ) \in k-\mathcal{ST}$$. Consequently, $$\psi ( z ) \in k-\mathcal{ST} ( \alpha ,\beta )$$ in $$\mathfrak{A.}$$ □

Similarly, we can prove that if $$f ( z ) \in k- \mathcal{ST}_{s}^{\ast } ( \alpha ,\beta )$$, then

$$\phi ( z ) =\frac{1}{2} \bigl[ f ( z ) -f ( -z ) \bigr] \in k- \mathcal{ST}^{\ast } ( \alpha ,\beta ) .$$

Taking $$\alpha =0$$, we obtain the following result proved by Noor .

### Corollary 3.2

Let $$f ( z ) \in k-\mathcal{ST}_{s}$$. Then the odd function

$$\psi ( z ) =\frac{1}{2} \bigl[ f ( z ) -f ( -z ) \bigr] \in k- \mathcal{ST}.$$

Note that, for $$k=\alpha =0$$, the function $$\psi ( z ) = \frac{1}{2} [ f ( z ) -f ( -z ) ]$$ is a starlike function in $$\mathfrak{A}$$; see .

### Theorem 3.3

Let $$\alpha \geq 2$$ and $$\beta >-1$$. Then $$k-\mathcal{ST} ( \alpha -1,\beta ) \subset k-\mathcal{ST} ( \alpha ,\beta )$$.

### Proof

Let $$f ( z ) \in k-\mathcal{ST} ( \alpha -1,\beta )$$ and set

$$p ( z ) =\frac{z ( \mathcal{I}_{\beta }^{\alpha }f ( z ) ) ^{\prime }}{\mathcal{I}_{\beta }^{\alpha }f ( z ) }.$$
(3.1)

Note that $$p ( z )$$ is analytic in $$\mathfrak{A}$$ with $$p ( 0 ) =1$$.

From (3.1) and identity (1.10) we have

$$\frac{\mathcal{I}_{\beta }^{\alpha -1}f ( z ) }{ \mathcal{I}_{\beta }^{\alpha }f ( z ) }= ( 1-\gamma ) p ( z ) +\gamma$$
(3.2)

with

$$\gamma =\frac{\beta }{\beta +1}.$$
(3.3)

Logarithmic differentiation of (3.2) yields

$$\frac{z ( \mathcal{I}_{\beta }^{\alpha -1}f ( z ) ) ^{\prime }}{\mathcal{I}_{\beta }^{\alpha -1}f ( z ) }= \biggl\{ p ( z ) +\frac{ ( 1-\gamma ) zp^{{\prime }} ( z ) }{ ( 1-\gamma ) zp ( z ) +\gamma } \biggr\} ,$$

and thus it follows that

$$\biggl( p ( z ) +\frac{zp^{{\prime }} ( z ) }{zp ( z ) +\beta } \biggr) \prec p_{k} ( z ) .$$

Using Lemma 2.1, we have

$$p ( z ) \prec q ( z ) \prec p_{k} ( z )$$

with

$$q ( z ) = \biggl[ \int _{0}^{1} \biggl( t^{\beta } \exp \int _{z}^{tz}\frac{p_{k} ( u ) -1}{u}\,du \biggr) \,dt \biggr] ^{-1}-\beta .$$

This proves that $$f ( z ) \in k-\mathcal{ST} ( \alpha ,\beta )$$ in $$\mathfrak{A}$$, and the proof is complete. □

### Theorem 3.4

Let $$\alpha \geq 2$$ and $$\beta >-1$$. Then $$k-\mathcal{ST}^{\ast } ( \alpha -1,\beta ) \subset k-\mathcal{ST}^{\ast } ( \alpha ,\beta )$$.

### Proof

Let

$$\frac{z ( \mathfrak{L}_{\beta }^{\alpha }f ( z ) ) ^{\prime }}{\mathfrak{L}_{\beta }^{\alpha }f ( z ) }=h ( z ) ,$$
(3.4)

where $$h ( z )$$ is analytic in $$\mathfrak{A}$$ with $$h ( 0 ) =1$$.

From (3.4) and identity (1.11) we get

$$\frac{1}{\alpha +\beta }\frac{z ( \mathfrak{L}_{\beta }^{\alpha }f ( z ) ) ^{\prime }}{\mathfrak{L}_{\beta }^{ \alpha }f ( z ) }+ \biggl( 1-\frac{1}{\alpha +\beta } \biggr) = \frac{\mathfrak{L}_{\beta }^{\alpha -1}f ( z ) }{ \mathfrak{L}_{\beta }^{\alpha }f ( z ) }.$$
(3.5)

Logarithmic differentiation of (3.5), together with (3.4), gives us

\begin{aligned} \frac{z ( \mathfrak{L}_{\beta }^{\alpha -1}f ( z ) ) ^{\prime }}{\mathfrak{L}_{\beta }^{\alpha -1}f ( z ) } =&h ( z ) +\frac{\frac{1}{\alpha +\beta }zh^{{\prime }} ( z ) }{\frac{1}{\alpha +\beta }h ( z ) +\frac{ \alpha +\beta -1}{\alpha +\beta }} \\ =&h ( z ) +\frac{zh^{{\prime }} ( z ) }{h ( z ) +\alpha +\beta -1}. \end{aligned}

Since $$f ( z ) \in k-\mathcal{ST}^{\ast } ( \alpha -1, \beta )$$, it follows that

$$h ( z ) +\frac{zh^{{\prime }} ( z ) }{h ( z ) +\alpha +\beta -1}\prec p_{k} ( z ) .$$

Applying Lemma 2.1, we have

$$h ( z ) \prec p_{k} ( z ) .$$

This proves our result. □

### Theorem 3.5

Let $$\alpha \geq 2$$ and $$\beta >-1$$. Then $$k-\mathcal{ST}_{s} ( \alpha -1,\beta ) \subset k-\mathcal{ST}_{s} ( \alpha , \beta )$$.

### Proof

Let $$f ( z ) \in k-\mathcal{ST}_{s} ( \alpha -1, \beta )$$. Then, using Theorems 3.1 and 3.3, we have

$$\psi ( z ) =\frac{f ( z ) -f ( -z ) }{2}\in k-\mathcal{ST} ( \alpha -1,\beta ) \subset k- \mathcal{ST} ( \alpha ,\beta ) .$$

From this it easily follows that $$f ( z ) \in k- \mathcal{ST}_{s} ( \alpha ,\beta )$$, and this completes the proof. □

A similar result for the class $$k-\mathcal{ST}_{s}^{\ast } ( \alpha ,\beta )$$ can be easily proved.

### Theorem 3.6

Let $$\alpha \geq 1$$ and $$\beta >0$$. Then $$k-\mathcal{UK}_{s} ( \alpha -1,\beta ) \subset k-\mathcal{UK}_{s} ( \alpha , \beta )$$.

### Proof

Let $$f ( z ) \in k-\mathcal{UK}_{s} ( \alpha -1, \beta )$$. Then there exists $$g ( z ) \in k- \mathcal{ST}_{s} ( \alpha -1,\beta )$$ such that

$$\frac{2z ( \mathcal{I}_{\beta }^{\alpha -1}f ( z ) ) ^{\prime }}{\mathcal{I}_{\beta }^{\alpha -1}g ( z ) -\mathcal{I}_{\beta }^{\alpha -1}g ( -z ) }=\frac{z ( \mathcal{I}_{\beta }^{\alpha -1}f ( z ) ) ^{\prime }}{\mathcal{I}_{\beta }^{\alpha -1}\psi ( z ) }\in P,$$

where $$\psi ( z ) =\frac{\mathcal{I}_{\beta }^{\alpha -1}g ( z ) -\mathcal{I}_{\beta }^{\alpha -1}g ( -z ) }{2}\in k-\mathcal{ST} ( \alpha -1,\beta ) \subset k- \mathcal{ST} ( \alpha ,\beta )$$ in $$\mathfrak{A}$$.

Let us set

$$\frac{z ( \mathcal{I}_{\beta }^{\alpha }f ( z ) ) ^{\prime }}{\mathcal{I}_{\beta }^{\alpha }\psi ( z ) }=p ( z ) ,$$
(3.6)

where $$p ( z )$$ is analytic in $$\mathfrak{A}$$ with $$p ( 0 ) =1$$. Then by (3.6) and identity (1.10) we get

$$\frac{\mathcal{I}_{\beta }^{\alpha -1}\psi ( z ) }{ \mathcal{I}_{\beta }^{\alpha }\psi ( z ) }= ( 1- \gamma ) p_{0} ( z ) +\gamma ,$$

where $$p_{0} ( z ) =\frac{z ( \mathcal{I}_{\beta } ^{\alpha }\psi ( z ) ) ^{\prime }}{\mathcal{I}_{ \beta }^{\alpha }\psi ( z ) }$$, and γ is given by (3.3). Now by simple computations we obtain

\begin{aligned} \frac{z ( \mathcal{I}_{\beta }^{\alpha -1}f ( z ) ) ^{\prime }}{z\mathcal{I}_{\beta }^{\alpha -1}\psi ( z ) } =&\frac{z ( \mathcal{I}_{\beta }^{\alpha -1}f ( z ) ) ^{\prime }}{\mathcal{I}_{\beta }^{\alpha }\psi ( z ) [ ( 1-\gamma ) p_{0} ( z ) +\gamma ] } \\ =&\frac{z [ ( z ( \mathcal{I}_{\beta }^{\alpha }f ( z ) ) ^{\prime } ) ^{\prime } ] + \beta z ( \mathcal{I}_{\beta }^{\alpha }f ( z ) ) ^{\prime }}{ ( \beta +1 ) \mathcal{I}_{\beta }^{\alpha } \psi ( z ) [ ( 1-\gamma ) p_{0} ( z ) +\gamma ] } \\ =&\frac{\beta p ( z ) +p ( z ) p_{0} ( z ) +zp^{\prime } ( z ) }{ ( \beta +1 ) [ ( 1-\frac{\beta }{1+\beta } ) p_{0} ( z ) +\frac{\beta }{1+\beta } ] } \\ =&\frac{\beta p ( z ) +p ( z ) p_{0} ( z ) +zp^{\prime } ( z ) }{p_{0} ( z ) + \beta } \\ =&p ( z ) +\frac{zp^{\prime } ( z ) }{p_{0} ( z ) +\beta }. \end{aligned}

Since $$f ( z ) \in k-\mathcal{UK}_{s} ( \alpha -1, \beta )$$, it follows that

$$p ( z ) +\frac{zp^{\prime } ( z ) }{p_{0} ( z ) +\beta }\in \mathcal{P}\quad \text{in }\mathfrak{A}.$$

Applying Lemma.2.2, we have $$p ( z ) \in \mathcal{P}$$ in $$\mathfrak{A}$$. This proves $$f ( z ) \in k-\mathcal{UK}_{s} ( \alpha ,\beta )$$ in $$\mathfrak{A}$$. □

By a similar argument we can easily prove the following inclusion result.

### Theorem 3.7

Let $$\alpha \geq 1$$ and $$\beta >0$$. Then $$k-\mathcal{UK}^{\ast } ( \alpha -1,\beta ) \subset k-\mathcal{UK}^{\ast } ( \alpha ,\beta )$$.

### Theorem 3.8

Let $$f ( z ) \in k-\mathcal{ST}_{s} ( \alpha ,\beta )$$ in $$\mathfrak{A}$$. Then

$$\mathfrak{Re} \biggl\{ \frac{z ( \mathcal{I}_{\beta }^{\alpha -1}f ( z ) ) ^{\prime }}{\mathcal{I}_{\beta }^{\alpha -1} \varphi ( z ) } \biggr\} >0$$

for $$\vert z \vert < R ( \beta ,\gamma _{0} )$$, where

$$R ( \beta ,\gamma _{0} ) =\frac{ ( 1+\beta ) }{ ( 2-\gamma _{0} ) +\sqrt{ ( 2-\gamma _{0} ) ^{2}+ ( 1+\beta ) ( \beta +2\gamma _{0}-1 ) }}$$

with

$$\gamma _{0}=\frac{k}{k+1}.$$
(3.7)

### Proof

Let $$f ( z ) \in k-\mathcal{ST}_{s} ( \alpha ,\beta )$$. Then

$$\varphi ( z ) =\frac{f ( z ) -f ( -z ) }{2}\in k-\mathcal{ST} ( \alpha ,\beta ) ,$$

and hence

$$\frac{z ( \mathcal{I}_{\beta }^{\alpha }f ( z ) ) ^{\prime }}{\mathcal{I}_{\beta }^{\alpha }\varphi ( z ) } \in \mathcal{P} ( p_{k} ) \subset \mathcal{P} ( \gamma _{0} ) ,$$

where $$\gamma _{0}$$ is given by (3.7). Let

\begin{aligned} \frac{z ( \mathcal{I}_{\beta }^{\alpha }f ( z ) ) ^{\prime }}{\mathcal{I}_{\beta }^{\alpha }\varphi ( z ) } =&h ( z ) ,\quad h ( z ) \in \mathcal{P} ( \gamma _{0} ) , \\ =& ( 1-\gamma _{0} ) h_{0} ( z ) +\gamma _{0},\quad h_{0} ( z ) \in \mathcal{P}. \end{aligned}
(3.8)

Then, proceeding as in Theorem 3.5, we have

$$\frac{z ( \mathcal{I}_{\beta }^{\alpha -1}f ( z ) ) ^{\prime }}{\mathcal{I}_{\beta }^{\alpha -1}\varphi ( z ) }=h ( z ) +\frac{zh^{\prime } ( z ) }{p ( z ) +\beta },$$
(3.9)

where $$p ( z ) =\frac{z ( \mathcal{I}_{\beta }^{ \alpha }\varphi ( z ) ) ^{\prime }}{\mathcal{I} _{\beta }^{\alpha }\varphi ( z ) }\in \mathcal{P} ( \gamma )$$. Using (3.8) and $$p ( z ) = ( 1- \gamma _{0} ) p_{0} ( z ) +\gamma _{0}$$ in (3.9), we have

$$\frac{z ( \mathcal{I}_{\beta }^{\alpha -1}f ( z ) ) ^{\prime }}{\mathcal{I}_{\beta }^{\alpha -1}\varphi ( z ) }= ( 1-\gamma _{0} ) h_{0} ( z ) + \gamma _{0}+\frac{ ( 1-\gamma _{0} ) zh_{0}^{\prime } ( z ) }{ ( 1-\gamma _{0} ) p_{0} ( z ) + \gamma _{0}+\beta }$$

with $$h_{0} ( z ) \in \mathcal{P}$$, $$p_{0} ( z ) \in \mathcal{P}$$, that is,

$$\frac{1}{1-\gamma _{0}} \biggl[ \frac{z ( \mathcal{I}_{\beta }^{ \alpha -1}f ( z ) ) ^{\prime }}{\mathcal{I}_{\beta }^{\alpha -1}\varphi ( z ) }-\gamma _{0} \biggr] =h_{0} ( z ) +\frac{zh_{0}^{\prime } ( z ) }{ ( 1- \gamma _{0} ) p_{0} ( z ) +\gamma _{0}+\beta }.$$

Using the distortion result for the class $$\mathcal{P}$$, we obtain

\begin{aligned} &\mathfrak{Re} \biggl[ \frac{1}{1-\gamma _{0}} \biggl\{ \frac{z ( \mathcal{I}_{\beta }^{\alpha -1}f ( z ) ) ^{\prime }}{\mathcal{I}_{\beta }^{\alpha -1}\varphi ( z ) }-\gamma _{0} \biggr\} \biggr] \\ &\quad \geq \mathfrak{Re}h_{0} ( z ) \biggl\{ 1-\frac{\frac{2r}{1-r^{2}}}{ ( 1-\gamma _{0} ) \frac{1-r}{1+r}+ ( \gamma _{0}+\beta ) } \biggr\} \\ &\quad =\mathfrak{Re}h_{0} ( z ) \biggl\{ 1-\frac{2r}{ ( 1- \gamma _{0} ) ( 1+r ) ^{2}+ ( 1-r^{2} ) ( \gamma _{0}+\beta ) } \biggr\} . \end{aligned}
(3.10)

Right-hand side of (3.10) is greater than or equal to zero for $$\vert z \vert < R ( \beta , \gamma _{0} )$$, where $$R ( \beta ,\gamma _{0} )$$ is the least positive root of the equation

$$T ( r ) := ( 1-\beta -2\gamma _{0} ) r^{2}-2 ( 2-\gamma _{0} ) r+ ( 1+\beta ) =0,$$

that is,

\begin{aligned} R ( \beta ,\gamma _{0} ) =&\frac{2 ( 2-\gamma _{0} ) -\sqrt{4 ( 2-\gamma _{0} ) ^{2}+4 ( 1+\beta ) ( \beta +2\gamma _{0}-1 ) }}{2 ( 1-\beta -2\gamma _{0} ) } \\ =&\frac{ ( 1+\beta ) }{ ( 2-\gamma _{0} ) +\sqrt{ ( 2-\gamma _{0} ) ^{2}+ ( 1+\beta ) ( \beta +2\gamma _{0}-1 ) }}. \end{aligned}

The proof is completed. □

Particular Cases

1. (i)

For $$\beta =0$$ and $$\gamma _{0}=\frac{k}{k+1}=0$$ (i.e., $$k=0$$), we have $$f ( z ) \in \mathcal{S}_{s}^{\ast } ( \alpha ,0 )$$ ($$\psi \in \mathcal{S}^{\ast } ( \alpha ,0 )$$) and

$$R ( 0,0 ) =\frac{1}{2+\sqrt{3}}.$$
2. (ii)

For $$k=1$$ and $$\beta =0$$,

$$R \biggl( 0,\frac{1}{2} \biggr) =\frac{1}{3}.$$
3. (iii)

For $$k=1$$ and $$\beta =1$$,

$$R \biggl( 1,\frac{1}{2} \biggr) =\frac{4}{4+\sqrt{17}}.$$

### Theorem 3.9

Let $$\mathfrak{L}_{\beta }^{\alpha }f ( z ) \in k- \mathcal{ST}$$. Then

$$\mathfrak{L}_{\beta }^{\alpha -1}f ( z ) \in \mathcal{S} ^{\ast } ( \gamma _{0} ) , \quad \gamma _{0}=\frac{k}{k+1}$$

for $$\vert z \vert < R_{1}$$, where

$$R_{1} ( \alpha ,\beta ,\gamma _{0} ) =\frac{\alpha +\beta }{2- \gamma _{0}+\sqrt{ ( 2-\gamma _{0} ) ^{2}+ ( \alpha + \beta ) ( 2\gamma _{0}+\alpha +\beta -2 ) }}.$$

### Proof

Since $$\mathfrak{L}_{\beta }^{\alpha }f ( z ) \in k- \mathcal{ST}$$, we have

$$\frac{z ( \mathfrak{L}_{\beta }^{\alpha }f ( z ) ) ^{\prime }}{\mathfrak{L}_{\beta }^{\alpha }f ( z ) }=h ( z ) ,\quad h ( z ) \prec p_{k} ( z )$$

in $$\mathfrak{A}$$. With a similar argument as in Theorem 3.5, we have

$$\frac{z ( \mathfrak{L}_{\beta }^{\alpha -1}f ( z ) ) ^{\prime }}{\mathfrak{L}_{\beta }^{\alpha -1}f ( z ) }=h ( z ) +\frac{zh^{\prime } ( z ) }{h ( z ) +\alpha +\beta -1},$$

that is,

\begin{aligned} &\mathfrak{Re} \biggl[ \frac{1}{1-\gamma _{0}} \biggl\{ \frac{z ( \mathfrak{L}_{\beta }^{\alpha -1}f ( z ) ) ^{\prime }}{\mathfrak{L}_{\beta }^{\alpha -1}f ( z ) }-\gamma _{0} \biggr\} \biggr] \\ &\quad =\mathfrak{Re} \biggl[ h_{0} ( z ) +\frac{zh_{0}^{\prime } ( z ) }{ ( 1-\gamma _{0} ) h_{0} ( z ) + ( \gamma _{0}+\alpha + \beta -1 ) } \biggr] \\ &\quad \geq \mathfrak{Re}h_{0} ( z ) \biggl[ 1-\frac{\frac{2r}{1-r ^{2}}}{ ( 1-\gamma _{0} ) \frac{1-r}{1+r}+ ( \gamma _{0}+\alpha +\beta -1 ) } \biggr] , \end{aligned}
(3.11)

where

$$h ( z ) = ( 1-\gamma _{0} ) h_{0} ( z ) +\gamma _{0},\quad h_{0}\in \mathcal{P}, \gamma _{0}= \frac{k}{k+1}.$$

The right-hand side of (3.11) is greater than or equal to zero for $$\vert z \vert < R_{1}$$, where $$R_{1}$$ is the least positive root of the equation

$$T ( r ) := ( 2-2\gamma _{0}-\alpha -\beta ) r ^{2}-2 ( 2- \gamma _{0} ) r+\alpha +\beta =0,$$

that is,

\begin{aligned} R_{1} ( \alpha ,\beta ,\gamma _{0} ) =&\frac{2-\gamma _{0}-\sqrt{ ( 2-\gamma _{0} ) ^{2}+ ( \alpha +\beta ) ( 2\gamma _{0}+\alpha +\beta -2 ) }}{2 ( 2-\alpha - \beta -2\gamma _{0} ) } \\ =&\frac{\alpha +\beta }{2-\gamma _{0}+\sqrt{ ( 2-\gamma _{0} ) ^{2}+ ( \alpha +\beta ) ( 2 \gamma _{0}+\alpha +\beta -2 ) }}. \end{aligned}

This completes the proof. □

## Conclusion

In this paper, we have defined some new classes of analytic functions involving integral operators. We have shown that these classes generalize the well-known classes, and already existing results can be obtained as a particular cases of our results. Inclusion relations of these classes are also a significant part of our work. We believe that the work presented in this paper will give researchers a new direction and will motivate them to explore more interesting facts on similar lines.

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### Acknowledgements

The authors would like to thank the reviewers of this paper for his/her valuable comments on the earlier version of the paper. They would also like to acknowledge Prof. Dr. Salim ur Rehman, V.C. Sarhad University of Science & I. T, for providing excellent research and academic environment.

Not applicable.

## Funding

Sarhad University of Science & I. T Peshawar.

## Author information

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### Contributions

All authors jointly worked on the results, and they read and approved the final manuscript.

### Corresponding author

Correspondence to Shahid Mahmood.

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### Competing interests

The authors declare that they have no competing interests. 