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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 [1].

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 [1].

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 [2] 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 [2] 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 [3] 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 [3] 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 [1]). 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 [12] 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 [13] 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

([4])

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

([18])

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 [12].

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 [2].

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.

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Mahmood, S., Khan, I., Srivastava, H.M. et al. Inclusion relations for certain families of integral operators associated with conic regions. J Inequal Appl 2019, 59 (2019). https://doi.org/10.1186/s13660-019-2015-9

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MSC

  • 30C45
  • 30C50

Keywords

  • Sakaguchi functions
  • Schwarz function
  • Subordination
  • Functions with positive real parts
  • Analytic functions
  • Conic domain
  • Uniformly starlike
  • Integral operators
  • Symmetrical points