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Applications of differential subordinations for certain classes of p-valent functions associated with generalized Srivastava-Attiya operator

Abstract

The object of the present paper is to investigate some inclusion relations and other interesting properties for certain classes of p-valent functions involving generalized Srivastava-Attiya operator by using the principle of differential subordination.

MSC:30C45.

1 Introduction

Let $A\left(p\right)$ be the class of functions which are analytic and p-valent in the unit disc $U=\left\{zâˆˆ\mathbb{C}:|z|<1\right\}$ of the form

$f\left(z\right)={z}^{p}+\underset{k=1}{\overset{\mathrm{âˆž}}{âˆ‘}}{a}_{k+p}{z}^{k+p}\phantom{\rule{1em}{0ex}}\left(pâˆˆ\mathbb{N}=\left\{1,2,â€¦\right\}\right).$
(1.1)

Let also $A\left(1\right)={A}_{1}$. For $g\left(z\right)âˆˆA\left(p\right)$, given by $g\left(z\right)={z}^{p}+{âˆ‘}_{k=1}^{\mathrm{âˆž}}{b}_{k+p}{z}^{k+p}$, the Hadamard product (or convolution) of $f\left(z\right)$ and $g\left(z\right)$ is defined by

$\left(fâˆ—g\right)\left(z\right)={z}^{p}+\underset{k=1}{\overset{\mathrm{âˆž}}{âˆ‘}}{a}_{k+p}{b}_{k+p}{z}^{k+p}=\left(gâˆ—f\right)\left(z\right).$
(1.2)

Next, in the usual notation, let $\mathrm{Î¦}\left(z,s,a\right)$ denote the Hurwitz-Lerch Zeta function defined as follows:

(1.3)

For further interesting properties and characteristics of the Hurwitz-Lerch Zeta function $\mathrm{Î¦}\left(z,s,a\right)$ see [2, 5, 8, 9, 11], and [21].

Recently, Srivastava and Attiya [20] have introduced the linear operator ${L}_{s,b}:{A}_{1}â†’{A}_{1}$, defined in terms of the Hadamard product by

${L}_{s,b}\left(f\right)\left(z\right)={G}_{s,b}\left(z\right)âˆ—f\left(z\right)\phantom{\rule{1em}{0ex}}\left(zâˆˆU;bâˆˆ\mathbb{C}\mathrm{âˆ–}{\mathbb{Z}}_{0}^{âˆ’};sâˆˆ\mathbb{C}\right),$
(1.4)

where

${G}_{s,b}={\left(1+b\right)}^{s}\left[\mathrm{Î¦}\left(z,s,b\right)âˆ’{b}^{âˆ’s}\right]\phantom{\rule{1em}{0ex}}\left(zâˆˆU\right).$
(1.5)

The Srivastava-Attiya operator ${L}_{s,b}$ contains, among its special cases, the integral operators introduced and investigated by Alexander [1], Libera [7] and Jung et al. [6].

Analogous to ${L}_{s,b}$, Liu [10] defined the operator ${J}_{p,s,b}:A\left(p\right)â†’A\left(p\right)$ by

${J}_{p,s,b}\left(f\right)\left(z\right)={G}_{p,s,b}\left(z\right)âˆ—f\left(z\right)\phantom{\rule{1em}{0ex}}\left(zâˆˆU;bâˆˆ\mathbb{C}\mathrm{âˆ–}{\mathbb{Z}}_{0}^{âˆ’};sâˆˆ\mathbb{C};pâˆˆ\mathbb{N}\right),$
(1.6)

where

${G}_{p,s,b}={\left(1+b\right)}^{s}\left[{\mathrm{Î¦}}_{p}\left(z,s,b\right)âˆ’{b}^{âˆ’s}\right]$

and

${\mathrm{Î¦}}_{p}\left(z,s,b\right)=\frac{1}{{b}^{s}}+\underset{k=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{z}^{k+p}}{{\left(k+1+b\right)}^{s}}.$
(1.7)

It is easy to observe from (1.6) and (1.7) that

${J}_{p,s,b}\left(f\right)\left(z\right)={z}^{p}+\underset{k=1}{\overset{\mathrm{âˆž}}{âˆ‘}}{\left(\frac{1+b}{k+1+b}\right)}^{s}{a}_{k+p}{z}^{k+p}.$
(1.8)

We note that

1. (i)

${J}_{p,0,b}\left(f\right)\left(z\right)=f\left(z\right)$;

2. (ii)

${J}_{1,1,0}\left(f\right)\left(z\right)=Lf\left(z\right)={âˆ«}_{0}^{z}\frac{f\left(t\right)}{t}\phantom{\rule{0.2em}{0ex}}dt$ ($fâˆˆ{A}_{1}$), where the operator L was introduced by Alexander [1];

3. (iii)

${J}_{1,s,b}\left(f\right)\left(z\right)={L}_{s,b}f\left(z\right)$ ($sâˆˆ\mathbb{C}$$bâˆˆ\mathbb{C}\mathrm{âˆ–}{\mathbb{Z}}_{0}^{âˆ’}$), where the operator ${L}_{s,b}$ was introduced by Srivastava-Attiya [20];

4. (iv)

${J}_{p,1,\mathrm{Î¼}+pâˆ’1}\left(f\right)\left(z\right)={F}_{\mathrm{Î¼},p}\left(f\right)\left(z\right)$ ($\mathrm{Î¼}>âˆ’p$$pâˆˆ\mathbb{N}$), where the operator ${F}_{\mathrm{Î¼},p}$ was introduced by Choi et al. [3];

5. (v)

${J}_{p,\mathrm{Î±},p}\left(f\right)\left(z\right)={I}_{p}^{\mathrm{Î±}}f\left(z\right)$ ($\mathrm{Î±}>0$$pâˆˆ\mathbb{N}$), where the operator ${I}_{p}^{\mathrm{Î±}}$ was introduced by Shams et al. [18];

6. (vi)

${J}_{p,\mathrm{Î³},pâˆ’1}\left(f\right)\left(z\right)={J}_{p}^{\mathrm{Î³}}f\left(z\right)$ ($\mathrm{Î³}âˆˆ{\mathbb{N}}_{0}=\mathbb{N}âˆª\left\{0\right\}$$pâˆˆ\mathbb{N}$), where the operator ${J}_{p}^{\mathrm{Î³}}$ was introduced by El-Ashwah and Aouf [4];

7. (vii)

${J}_{p,\mathrm{Î³},p+lâˆ’1}\left(f\right)\left(z\right)={J}_{p}^{\mathrm{Î³}}\left(l\right)f\left(z\right)$ ($\mathrm{Î³}âˆˆ{\mathbb{N}}_{0}$$pâˆˆ\mathbb{N}$$lâ‰¥0$), where the operator ${J}_{p}^{\mathrm{Î³}}\left(l\right)$ was introduced by El-Ashwah and Aouf [4].

It follows from (1.8) that

$z{\left({J}_{p,s,b}\left(f\right)\left(z\right)\right)}^{\mathrm{â€²}}=\left(b+1\right){J}_{p,sâˆ’1,b}\left(f\right)\left(z\right)âˆ’\left(b+1âˆ’p\right){J}_{p,s,b}\left(f\right)\left(z\right).$
(1.9)

For two analytic functions $f,gâˆˆA\left(p\right)$, we say that f is subordinate to g, written $f\left(z\right)â‰ºg\left(z\right)$ if there exists a Schwarz function $w\left(z\right)$, which (by definition) is analytic in U with $w\left(0\right)=0$ and $|w\left(z\right)|<1$ for all $zâˆˆU$, such that $f\left(z\right)=g\left(w\left(z\right)\right)$$zâˆˆU$. Furthermore, if the function $g\left(z\right)$ is univalent in U, then we have the following equivalence (see [14]):

$f\left(z\right)â‰ºg\left(z\right)\phantom{\rule{1em}{0ex}}â‡”\phantom{\rule{1em}{0ex}}f\left(0\right)=g\left(0\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}f\left(U\right)âŠ‚g\left(U\right).$

Definition 1 For fixed parameters A and B, with $âˆ’1â‰¤B, we say that $fâˆˆA\left(p\right)$ is in the class ${S}_{p}^{s,b}\left(A,B\right)$ if it satisfies the following subordination condition:

$\frac{{\left({J}_{p,s,b}\left(f\right)\left(z\right)\right)}^{\mathrm{â€²}}}{p{z}^{pâˆ’1}}â‰º\frac{1+Az}{1+Bz}\phantom{\rule{1em}{0ex}}\left(pâˆˆ\mathbb{N}\right).$
(1.10)

In view of the definition of subordination (1.10) is equivalent to the following condition:

$|\frac{\frac{{\left({J}_{p,s,b}\left(f\right)\left(z\right)\right)}^{\mathrm{â€²}}}{p{z}^{pâˆ’1}}âˆ’1}{B\frac{{\left({J}_{p,s,b}\left(f\right)\left(z\right)\right)}^{\mathrm{â€²}}}{p{z}^{pâˆ’1}}âˆ’A}|<1\phantom{\rule{1em}{0ex}}\left(zâˆˆU\right).$

For convenience, we write ${S}_{p}^{s,b}\left(1âˆ’\frac{2\mathrm{Î·}}{p},âˆ’1\right)={S}_{p}^{s,b}\left(\mathrm{Î·}\right)$, where ${S}_{p}^{s,b}\left(\mathrm{Î·}\right)$ denotes the class of functions in $A\left(p\right)$ satisfying the inequality

$Re\left(\frac{{\left({J}_{p,s,b}\left(f\right)\left(z\right)\right)}^{\mathrm{â€²}}}{p{z}^{pâˆ’1}}\right)>\mathrm{Î·}\phantom{\rule{1em}{0ex}}\left(0â‰¤\mathrm{Î·}<1;pâˆˆ\mathbb{N};zâˆˆU\right).$

In the present paper, we investigate some inclusion relations and other interesting properties for certain classes of p-valent functions involving an integral operator.

2 Preliminaries

To establish our main results, we need the following lemmas.

Lemma 1 ([13, 14])

Let h be analytic and convex (univalent) in U with$h\left(0\right)=1$. Suppose also that the function Ï† given by

$\mathrm{Ï†}\left(z\right)=1+{c}_{m}{z}^{m}+{c}_{m+1}{z}^{m+1}+â‹¯,$
(2.1)

is analytic in U, where m is a positive integer. If

(2.2)

then

$\mathrm{Ï†}\left(z\right)â‰º\mathrm{Ïˆ}\left(z\right)=\frac{\mathrm{Ï±}}{m}{z}^{âˆ’\frac{\mathrm{Ï±}}{m}}{âˆ«}_{0}^{z}{t}^{\frac{\mathrm{Ï±}}{m}âˆ’1}h\left(t\right)\phantom{\rule{0.2em}{0ex}}dtâ‰ºh\left(z\right)$
(2.3)

and$\mathrm{Ïˆ}\left(z\right)$is the best dominant of (2.2).

We denote by $H\left(\mathrm{Ï±}\right)$ the class of functions $\mathrm{Î¦}\left(z\right)$ given by

$\mathrm{Î¦}\left(z\right)=1+{c}_{1}z+{c}_{2}{z}^{2}+â‹¯,$
(2.4)

which are analytic in U and satisfy the following inequality:

$Re\left\{\mathrm{Î¦}\left(z\right)\right\}>\mathrm{Ï±}\phantom{\rule{1em}{0ex}}\left(0â‰¤\mathrm{Ï±}<1;zâˆˆU\right).$

Lemma 2 ([17])

Let the function$\mathrm{Î¦}\left(z\right)âˆˆH\left(\mathrm{Ï±}\right)$, where$\mathrm{Î¦}\left(z\right)$given by (2.4). Then

$Re\left\{\mathrm{Î¦}\left(\mathrm{Ï±}\right)\right\}â‰¥2\mathrm{Ï±}âˆ’1+\frac{2\left(1âˆ’\mathrm{Ï±}\right)}{1+|z|}\phantom{\rule{1em}{0ex}}\left(0â‰¤\mathrm{Ï±}<1;zâˆˆU\right).$

Lemma 3 ([22])

For$0â‰¤{\mathrm{Ï±}}_{1}$, ${\mathrm{Ï±}}_{2}<1$,

$H\left({\mathrm{Ï±}}_{1}\right)âˆ—H\left({\mathrm{Ï±}}_{2}\right)âŠ‚H\left({\mathrm{Ï±}}_{3}\right),\phantom{\rule{1em}{0ex}}{\mathrm{Ï±}}_{3}=1âˆ’2\left(1âˆ’{\mathrm{Ï±}}_{1}\right)\left(1âˆ’{\mathrm{Ï±}}_{2}\right).$

The result is best possible.

Lemma 4 ([24])

Let Î¼ be a positive measure on the unit interval$\left[0,1\right]$. Let$g\left(z,t\right)$be a complex valued function defined on$UÃ—\left[0,1\right]$such that$g\left(0,t\right)$is analytic in U for each$tâˆˆ\left[0,1\right]$and such that$g\left(z,0\right)$is Î¼ integrable on$\left[0,1\right]$for all$zâˆˆU$. In addition, suppose that$Re\left\{g\left(z,t\right)\right\}>0$, $g\left(âˆ’r,t\right)$is real and

$Re\left\{\frac{1}{g\left(z,t\right)}\right\}â‰¥\frac{1}{g\left(âˆ’r,t\right)}\phantom{\rule{1em}{0ex}}\left(|z|â‰¤r<1;tâˆˆ\left[0,1\right]\right).$

If G is defined by

$G\left(z\right)={âˆ«}_{0}^{1}g\left(z,t\right)\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¼}\left(t\right),$

then

$Re\left\{\frac{1}{G\left(z\right)}\right\}â‰¥\frac{1}{G\left(âˆ’r\right)}\phantom{\rule{1em}{0ex}}\left(|z|â‰¤r<1\right).$

Lemma 5 ([19])

Let the function g be analytic in U with$g\left(0\right)=1$and$Re\left\{g\left(z\right)\right\}>\frac{1}{2}$ ($zâˆˆU$). Then, for any function F analytic in U, $\left(gâˆ—F\right)\left(U\right)$is contained in the convex hull of$F\left(U\right)$.

Lemma 6 ([16])

Let Ï† be analytic in U with$\mathrm{Ï†}\left(0\right)=1$and$\mathrm{Ï†}\left(z\right)=0$for$0<|z|<1$and let$A,Bâˆˆ\mathbb{C}$with, $|B|â‰¤1$.

1. (i)

Letand$\mathrm{Î³}âˆˆ{\mathbb{C}}^{âˆ—}=\mathbb{C}\mathrm{âˆ–}\left\{0\right\}$satisfy either$|\frac{\mathrm{Î³}\left(Aâˆ’B\right)}{B}âˆ’1|â‰¤1$or$|\frac{\mathrm{Î³}\left(Aâˆ’B\right)}{B}+1|â‰¤1$. If Ï† satisfies

$1+\frac{z{\mathrm{Ï†}}^{\mathrm{â€²}}\left(z\right)}{\mathrm{Î³}\mathrm{Ï†}\left(z\right)}â‰º\frac{1+Az}{1+Bz},$
(2.5)

then

$\mathrm{Ï†}\left(z\right)â‰º{\left(1+Bz\right)}^{\mathrm{Î³}\left(\frac{Aâˆ’B}{B}\right)},$

and this is best dominant.

1. (ii)

Let$B=0$and$\mathrm{Î³}âˆˆ{\mathbb{C}}^{âˆ—}$be such that$|\mathrm{Î³}A|<\mathrm{Ï€}$. If Ï† satisfies (2.5), then

$\mathrm{Ï†}\left(z\right)â‰º{e}^{\mathrm{Î³}Az}$

and this is the best dominant.

For real or complex numbers a n and c ($câˆ‰{\mathbb{Z}}_{0}^{âˆ’}$) and $zâˆˆU$, the Gaussian hypergeometric function defined by

$\begin{array}{rl}{}_{2}F_{1}\left(a,n;c;z\right)& =1+\frac{an}{c}â‹\dots \frac{z}{1!}+\frac{a\left(a+1\right)n\left(n+1\right)}{c\left(c+1\right)}â‹\dots \frac{{z}^{2}}{2!}+â‹¯\\ =\underset{k=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{{\left(a\right)}_{k}{\left(n\right)}_{k}}{{\left(c\right)}_{k}}\frac{{z}^{k}}{k!},\end{array}$
(2.6)

where ${\left(d\right)}_{k}=d\left(d+1\right)â‹¯\left(d+kâˆ’1\right)$ and ${\left(d\right)}_{0}=1$. We note that the series defined by (2.6) converges absolutely for $zâˆˆU$, and hence, ${}_{2}F_{1}$ represents an analytic function in U (see, for details, [23], Ch.14]).

Lemma 7 ([23])

For real or complex numbers a, n and c ($câˆ‰{\mathbb{Z}}_{0}^{âˆ’}$)

(2.7)
(2.8)

and

${}_{2}F_{1}\left(a,n;c;z\right){=}_{2}{F}_{1}\left(n,a;c;z\right).$
(2.9)

3 Main results

Unless otherwise mentioned, we assume throughout this paper that $âˆ’1â‰¤B, $sâˆˆ\mathbb{C}$, $bâˆˆ\mathbb{C}\mathrm{âˆ–}{\mathbb{Z}}_{0}^{âˆ’}$, $pâˆˆ\mathbb{N}\mathrm{âˆ–}\left\{1\right\}$, $0<\mathrm{Î±}â‰¤1$, m is a positive integer and the powers are understood as principle values.

Theorem 1 Let f given by (1.1) satisfy the following subordination condition:

$\left(1âˆ’\mathrm{Î±}\right)\frac{{\left({J}_{p,s,b}\left(f\right)\left(z\right)\right)}^{\mathrm{â€²}}}{p{z}^{pâˆ’1}}+\mathrm{Î±}\frac{{\left({J}_{p,s,b}\left(f\right)\left(z\right)\right)}^{\mathrm{â€²}\mathrm{â€²}}}{p\left(pâˆ’1\right){z}^{pâˆ’2}}â‰º\frac{1+Az}{1+Bz}.$
(3.1)

Then

$\frac{{\left({J}_{p,s,b}\left(f\right)\left(z\right)\right)}^{\mathrm{â€²}}}{p{z}^{pâˆ’1}}â‰º\mathrm{Î¨}\left(z\right)â‰º\frac{1+Az}{1+Bz},$
(3.2)

where

(3.3)

is the best dominant of (3.2). Furthermore,

$fâˆˆ{S}_{p}^{s,b}\left(\mathrm{Î²}\right),$
(3.4)

where

(3.5)

The estimate (3.4) is best possible.

Proof Let

$\mathrm{Î¸}\left(z\right)=\frac{{\left({J}_{p,s,b}\left(f\right)\left(z\right)\right)}^{\mathrm{â€²}}}{p{z}^{pâˆ’1}}\phantom{\rule{1em}{0ex}}\left(zâˆˆU\right),$
(3.6)

where Î¸ is of the form (2.1) and is analytic in U. Differentiating (3.6) with respect to z, we get

$\left(1âˆ’\mathrm{Î±}\right)\frac{{\left({J}_{p,s,b}\left(f\right)\left(z\right)\right)}^{\mathrm{â€²}}}{p{z}^{pâˆ’1}}+\mathrm{Î±}\frac{{\left({J}_{p,s,b}\left(f\right)\left(z\right)\right)}^{\mathrm{â€²}\mathrm{â€²}}}{p\left(pâˆ’1\right){z}^{pâˆ’2}}=\mathrm{Î¸}\left(z\right)+\frac{\mathrm{Î±}}{pâˆ’1}z{\mathrm{Î¸}}^{\mathrm{â€²}}\left(z\right)â‰º\frac{1+Az}{1+Bz}.$

Applying Lemma 1 for $\mathrm{Ï±}=\frac{pâˆ’1}{\mathrm{Î±}}$ and Lemma 7, we have

This proves the assertion (3.2) of Theorem 1. Next, in order to prove the assertion (3.4) of Theorem 1, it suffices to show that

$\underset{|z|<1}{inf}\left\{Re\left(\mathrm{Î¨}\left(z\right)\right)\right\}=\mathrm{Î¨}\left(âˆ’1\right).$

Indeed, we have

$Re\left\{\frac{1+Az}{1+Bz}\right\}â‰¥\frac{1âˆ’Ar}{1âˆ’Br}\phantom{\rule{1em}{0ex}}\left(|z|â‰¤r<1\right).$

Setting

$G\left(z,\mathrm{Î¶}\right)=\frac{1+A\mathrm{Î¶}z}{1+B\mathrm{Î¶}z}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}d\mathrm{Î½}\left(\mathrm{Î¶}\right)=\frac{pâˆ’1}{m\mathrm{Î±}}{\mathrm{Î¶}}^{\frac{pâˆ’1}{m\mathrm{Î±}}âˆ’1}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î¶}\phantom{\rule{1em}{0ex}}\left(0â‰¤\mathrm{Î¶}â‰¤1\right),$

which is a positive measure on the closed interval $\left[0,1\right]$, we get

$\mathrm{Î¨}\left(z\right)={âˆ«}_{0}^{1}G\left(z,\mathrm{Î¶}\right)\phantom{\rule{0.2em}{0ex}}d\mathrm{Î½}\left(\mathrm{Î¶}\right).$

Then

$Re\left\{\mathrm{Î¨}\left(z\right)\right\}â‰¥{âˆ«}_{0}^{1}\frac{1âˆ’A\mathrm{Î¶}r}{1âˆ’B\mathrm{Î¶}r}\phantom{\rule{0.2em}{0ex}}d\mathrm{Î½}\left(\mathrm{Î¶}\right)=\mathrm{Î¨}\left(âˆ’r\right)\phantom{\rule{1em}{0ex}}\left(|z|â‰¤r<1\right).$

Letting $râ†’{1}^{âˆ’}$ in the above inequality, we obtain the assertion (3.4). Finally, the estimate (3.4) is best possible as Î¨ is the best dominant of (3.2). This completes the proof of Theorem 1.â€ƒâ–¡

Theorem 2 If$fâˆˆ{S}_{p}^{s,b}\left(\mathrm{Î·}\right)$ ($0â‰¤\mathrm{Î·}<1$), then

$Re\left\{\left(1âˆ’\mathrm{Î±}\right)\frac{{\left({J}_{p,s,b}\left(f\right)\left(z\right)\right)}^{\mathrm{â€²}}}{p{z}^{pâˆ’1}}+\mathrm{Î±}\frac{{\left({J}_{p,s,b}\left(f\right)\left(z\right)\right)}^{\mathrm{â€²}\mathrm{â€²}}}{p\left(pâˆ’1\right){z}^{pâˆ’2}}\right\}>\mathrm{Î·}\phantom{\rule{1em}{0ex}}\left(|z|

where

$R={\left\{\frac{\sqrt{{\left(pâˆ’1\right)}^{2}+{\left(m\mathrm{Î±}\right)}^{2}}âˆ’m\mathrm{Î±}}{pâˆ’1}\right\}}^{\frac{1}{m}}.$
(3.7)

The result is best possible.

Proof Let $fâˆˆ{S}_{p}^{s,b}\left(\mathrm{Î·}\right)$, then we write

$\frac{{\left({J}_{p,s,b}\left(f\right)\left(z\right)\right)}^{\mathrm{â€²}}}{p{z}^{pâˆ’1}}=\mathrm{Î·}+\left(1âˆ’\mathrm{Î·}\right)u\left(z\right)\phantom{\rule{1em}{0ex}}\left(zâˆˆU\right),$
(3.8)

where u is of the form (2.1), is analytic in U and has a positive real part in U. Differentiating (3.8) with respect to z, we have

$\frac{1}{1âˆ’\mathrm{Î·}}\left\{\left(1âˆ’\mathrm{Î±}\right)\frac{{\left({J}_{p,s,b}\left(f\right)\left(z\right)\right)}^{\mathrm{â€²}}}{p{z}^{pâˆ’1}}+\mathrm{Î±}\frac{{\left({J}_{p,s,b}\left(f\right)\left(z\right)\right)}^{\mathrm{â€²}\mathrm{â€²}}}{p\left(pâˆ’1\right){z}^{pâˆ’2}}âˆ’\mathrm{Î·}\right\}=u\left(z\right)+\frac{\mathrm{Î±}}{pâˆ’1}z{u}^{\mathrm{â€²}}\left(z\right).$
(3.9)

Applying the following well-known estimate [12]:

$\frac{|z{u}^{\mathrm{â€²}}\left(z\right)|}{Re\left\{u\left(z\right)\right\}}â‰¤\frac{2m{r}^{m}}{1âˆ’{r}^{2m}}\phantom{\rule{1em}{0ex}}\left(|z|=r<1\right),$

in (3.9), we have

(3.10)

such that the right-hand side of (3.10) is positive, if $r, where R is given by (3.7).

In order to show that the bound R is best possible, we consider the function $fâˆˆA\left(p\right)$ defined by

$\frac{{\left({J}_{p,s,b}\left(f\right)\left(z\right)\right)}^{\mathrm{â€²}}}{p{z}^{pâˆ’1}}=\mathrm{Î·}+\left(1âˆ’\mathrm{Î·}\right)\frac{1+{z}^{m}}{1âˆ’{z}^{m}}\phantom{\rule{1em}{0ex}}\left(0â‰¤\mathrm{Î·}<1;zâˆˆU\right).$

Note that

$\frac{1}{1âˆ’\mathrm{Î·}}\left\{\left(1âˆ’\mathrm{Î±}\right)\frac{{\left({J}_{p,s,b}\left(f\right)\left(z\right)\right)}^{\mathrm{â€²}}}{p{z}^{pâˆ’1}}+\mathrm{Î±}\frac{{\left({J}_{p,s,b}\left(f\right)\left(z\right)\right)}^{\mathrm{â€²}\mathrm{â€²}}}{p\left(pâˆ’1\right){z}^{pâˆ’2}}âˆ’\mathrm{Î·}\right\}=\frac{\left(pâˆ’1\right)\left(1âˆ’{z}^{2m}\right)âˆ’2\mathrm{Î±}m{z}^{m}}{\left(pâˆ’1\right){\left(1âˆ’{z}^{m}\right)}^{2}}=0,$

for $z={R}^{â‹\dots }exp\left\{\frac{\mathbf{i}\mathbit{Ï€}}{\mathbf{m}}\right\}$. This completes the proof of Theorem 2.â€ƒâ–¡

For a function $fâˆˆA\left(p\right)$, the generalized Bernardi-Libera-Livingston integral operator ${F}_{\mathrm{Î¼},p}$ is defined by

$\begin{array}{rl}{F}_{\mathrm{Î¼},p}\left(f\right)\left(z\right)& =\frac{\mathrm{Î¼}+p}{{z}^{\mathrm{Î¼}}}{âˆ«}_{0}^{z}{t}^{\mathrm{Î¼}âˆ’1}f\left(t\right)\phantom{\rule{0.2em}{0ex}}dt=\left({z}^{p}+\underset{k=1}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{\mathrm{Î¼}+p}{\mathrm{Î¼}+p+k}{z}^{k+p}\right)âˆ—f\left(z\right)\\ =z^{p}{}_{2}{F}_{1}\left(1,\mathrm{Î¼}+p;\mathrm{Î¼}+p+1;z\right)âˆ—f\left(z\right)\phantom{\rule{1em}{0ex}}\left(\mathrm{Î¼}>âˆ’p;zâˆˆU\right).\end{array}$
(3.11)

From (1.8) and (3.11), we have

$z{\left({J}_{p,s,b}{\mathrm{F}}_{\mathrm{Î¼},p}\left(f\left(z\right)\right)\right)}^{\mathrm{â€²}}=\left(\mathrm{Î¼}+p\right){J}_{p,s,b}\left(f\right)\left(z\right)âˆ’\mathrm{Î¼}{J}_{p,s,b}{\mathrm{F}}_{\mathrm{Î¼},p}\left(f\left(z\right)\right)\phantom{\rule{1em}{0ex}}\left(\mathrm{Î¼}>âˆ’p;zâˆˆU\right)$
(3.12)

and

${J}_{p,s,b}{\mathrm{F}}_{\mathrm{Î¼},p}\left(f\left(z\right)\right)={\mathrm{F}}_{\mathrm{Î¼},p}\left({J}_{p,s,b}\left(f\right)\left(z\right)\right).$

Theorem 3 Let$fâˆˆ{S}_{p}^{s,b}\left(A,B\right)$and${F}_{\mathrm{Î¼},p}$be defined by (3.11). Then

$\frac{{\left({J}_{p,s,b}{\mathrm{F}}_{\mathrm{Î¼},p}\left(f\left(z\right)\right)\right)}^{\mathrm{â€²}}}{p{z}^{pâˆ’1}}â‰º\mathrm{Î¦}\left(z\right)â‰º\frac{1+Az}{1+Bz},$
(3.13)

where

(3.14)

is the best dominant of (3.13). Furthermore,

$Re\left\{\frac{{\left({J}_{p,s,b}{\mathrm{F}}_{\mathrm{Î¼},p}\left(f\left(z\right)\right)\right)}^{\mathrm{â€²}}}{p{z}^{pâˆ’1}}\right\}>\mathrm{Ïˆ}\phantom{\rule{1em}{0ex}}\left(zâˆˆU\right),$

where

The result is best possible.

Proof Let

$K\left(z\right)=\frac{{\left({J}_{p,s,b}{\mathrm{F}}_{\mathrm{Î¼},p}\left(f\left(z\right)\right)\right)}^{\mathrm{â€²}}}{p{z}^{pâˆ’1}}\phantom{\rule{1em}{0ex}}\left(zâˆˆU\right),$
(3.15)

where K is of the form (2.1) and is analytic in U. Using (3.12) in (3.15) and differentiating the resulting equation with respect to z, we have

$\frac{{\left({J}_{p,s,b}\left(f\right)\left(z\right)\right)}^{\mathrm{â€²}}}{p{z}^{pâˆ’1}}=K\left(z\right)+\frac{z{K}^{\mathrm{â€²}}\left(z\right)}{p+\mathrm{Î¼}}â‰º\frac{1+Az}{1+Bz}.$

The remaining part of the proof is similar to that of Theorem 1, and so we omit it.â€ƒâ–¡

We note that

$\frac{{\left({J}_{p,s,b}{\mathrm{F}}_{\mathrm{Î¼},p}\left(f\left(z\right)\right)\right)}^{\mathrm{â€²}}}{p{z}^{pâˆ’1}}=\frac{p+\mathrm{Î¼}}{p{z}^{p+\mathrm{Î¼}}}{âˆ«}_{0}^{z}{t}^{\mathrm{Î¼}}{\left({J}_{p,s,b}\left(f\right)\left(t\right)\right)}^{\mathrm{â€²}}\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{1em}{0ex}}\left(fâˆˆA\left(p\right);zâˆˆU\right).$
(3.16)

Putting $A=1âˆ’\frac{2\mathrm{Î´}}{p}$ ($0â‰¤\mathrm{Î´}<1$) and $B=âˆ’1$ in Theorem 3 and using (3.16), we obtain the following corollary.

Corollary 1 If$fâˆˆA\left(p\right)$satisfies the following inequality:

$Re\left\{\frac{{\left({J}_{p,s,b}f\left(z\right)\right)}^{\mathrm{â€²}}}{p{z}^{pâˆ’1}}\right\}>\mathrm{Î´}\phantom{\rule{1em}{0ex}}\left(0â‰¤\mathrm{Î´}<1;zâˆˆU\right),$

then

$Re\left\{\frac{p+\mathrm{Î¼}}{p{z}^{p+\mathrm{Î¼}}}{âˆ«}_{0}^{z}{t}^{\mathrm{Î¼}}{\left({J}_{p,s,b}\left(f\right)\left(t\right)\right)}^{\mathrm{â€²}}\phantom{\rule{0.2em}{0ex}}dt\right\}>\frac{\mathrm{Î´}}{p}+\left(1âˆ’\frac{\mathrm{Î´}}{p}\right){\left[}_{2}{F}_{1}\left(1,1;\frac{\mathrm{Î¼}+p}{m}+1;\frac{1}{2}\right)âˆ’1\right]\phantom{\rule{1em}{0ex}}\left(zâˆˆU\right).$

The result is best possible.

Theorem 4 Let$f,gâˆˆA\left(p\right)$satisfy the following inequality:

$Re\left\{\frac{{J}_{p,s,b}\left(g\right)\left(z\right)}{{z}^{p}}\right\}>0\phantom{\rule{1em}{0ex}}\left(zâˆˆU\right).$

If

$|\frac{{J}_{p,s,b}\left(f\right)\left(z\right)}{{J}_{p,s,b}\left(g\right)\left(z\right)}âˆ’1|<1\phantom{\rule{1em}{0ex}}\left(zâˆˆU\right),$

then

where

(3.17)

Proof Let

$q\left(z\right)=\frac{{J}_{p,s,b}\left(f\right)\left(z\right)}{{J}_{p,s,b}\left(g\right)\left(z\right)}âˆ’1={c}_{m}{z}^{m}+{c}_{m+1}{z}^{m+1}+â‹¯,$
(3.18)

where $q\left(z\right)$ is analytic in U with $q\left(0\right)=0$ and $|q\left(z\right)|â‰¤|z{|}^{m}$. Then, by applying the familiar Schwartz Lemma [15], we have $q\left(z\right)={z}^{m}\mathcal{X}\left(z\right)$, where $\mathcal{X}$ is analytic in U and $|\mathcal{X}\left(z\right)|â‰¤1$. Therefore (3.18) leads to

${J}_{p,s,b}\left(f\right)\left(z\right)={J}_{p,s,b}\left(g\right)\left(z\right)\left(1+{z}^{m}\mathcal{X}\left(z\right)\right)\phantom{\rule{1em}{0ex}}\left(zâˆˆU\right).$
(3.19)

Differentiating (3.19) logarithmically with respect to z, we have

$\frac{z{\left({J}_{p,s,b}\left(f\right)\left(z\right)\right)}^{\mathrm{â€²}}}{{J}_{p,s,b}\left(f\right)\left(z\right)}=\frac{z{\left({J}_{p,s,b}\left(g\right)\left(z\right)\right)}^{\mathrm{â€²}}}{{J}_{p,s,b}\left(g\right)\left(z\right)}+\frac{{z}^{m}\left\{m\mathcal{X}\left(z\right)+z{\mathcal{X}}^{\mathrm{â€²}}\left(z\right)\right\}}{1+{z}^{m}\mathcal{X}\left(z\right)}.$
(3.20)

Letting

$\mathrm{Ï‰}\left(z\right)=\frac{{J}_{p,s,b}\left(g\right)\left(z\right)}{{z}^{p}}\phantom{\rule{1em}{0ex}}\left(zâˆˆU\right),$

where Ï‰ is in the form (2.1), is analytic in U$Re\left\{\mathrm{Ï‰}\left(z\right)\right\}>0$ and

$\frac{z{\left({J}_{p,s,b}\left(g\right)\left(z\right)\right)}^{\mathrm{â€²}}}{{J}_{p,s,b}\left(g\right)\left(z\right)}=\frac{z{\mathrm{Ï‰}}^{\mathrm{â€²}}\left(z\right)}{\mathrm{Ï‰}\left(z\right)}+p,$

then we have

$Re\left\{\frac{z{\left({J}_{p,s,b}\left(f\right)\left(z\right)\right)}^{\mathrm{â€²}}}{{J}_{p,s,b}\left(f\right)\left(z\right)}\right\}â‰¥pâˆ’|\frac{z{\mathrm{Ï‰}}^{\mathrm{â€²}}\left(z\right)}{\mathrm{Ï‰}\left(z\right)}|âˆ’|\frac{{z}^{m}\left\{m\mathcal{X}\left(z\right)+z{\mathcal{X}}^{\mathrm{â€²}}\left(z\right)\right\}}{1+{z}^{m}\mathcal{X}\left(z\right)}|.$
(3.21)

Using the following known estimates [12] (see also [15]):

$|\frac{{\mathrm{Ï‰}}^{\mathrm{â€²}}\left(z\right)}{\mathrm{Ï‰}\left(z\right)}|â‰¤\frac{2m{r}^{mâˆ’1}}{1âˆ’{r}^{2m}}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}|\frac{m\mathcal{X}\left(z\right)+z{\mathcal{X}}^{\mathrm{â€²}}\left(z\right)}{1+{z}^{m}\mathcal{X}\left(z\right)}|â‰¤\frac{m}{1âˆ’{r}^{m}}\phantom{\rule{1em}{0ex}}\left(|z|=r<1\right),$

in (3.21), we have

$Re\left\{\frac{z{\left({J}_{p,s,b}\left(f\right)\left(z\right)\right)}^{\mathrm{â€²}}}{{J}_{p,s,b}\left(f\right)\left(z\right)}\right\}â‰¥\frac{pâˆ’3m{r}^{m}âˆ’\left(p+m\right){r}^{2m}}{1âˆ’{r}^{2m}}\phantom{\rule{1em}{0ex}}\left(|z|=r<1\right),$

which is certainly positive, provided that , where is given by (3.17). This completes the proof of Theorem 4.â€ƒâ–¡

Theorem 5 Let$âˆ’1â‰¤{B}_{i}<{A}_{i}â‰¤1$ ($i=1,2$) and$\mathrm{Ï„}. If each of the functions${f}_{i}âˆˆA\left(p\right)$satisfies the following subordination condition:

$\left(1âˆ’\mathrm{Î±}\right)\frac{{\left({J}_{p,s,b}\left({f}_{i}\right)\left(z\right)\right)}^{\mathrm{â€²}}}{p{z}^{pâˆ’1}}+\mathrm{Î±}\frac{{\left({J}_{p,s,b}\left({f}_{i}\right)\left(z\right)\right)}^{\mathrm{â€²}\mathrm{â€²}}}{p\left(pâˆ’1\right){z}^{pâˆ’2}}â‰º\frac{1+{A}_{i}z}{1+{B}_{i}z}\phantom{\rule{1em}{0ex}}\left(i=1,2\right),$
(3.22)

then

$\left(1âˆ’\mathrm{Î±}\right)\frac{{\left({J}_{p,s,b}\left(F\right)\left(z\right)\right)}^{\mathrm{â€²}}}{p{z}^{pâˆ’1}}+\mathrm{Î±}\frac{{\left({J}_{p,s,b}\left(F\right)\left(z\right)\right)}^{\mathrm{â€²}\mathrm{â€²}}}{p\left(pâˆ’1\right){z}^{pâˆ’2}}â‰º\frac{1+\left(1âˆ’\frac{2\mathrm{Ï„}}{p}\right)z}{1âˆ’z},$
(3.23)

where

$F\left(z\right)={J}_{p,s,b}\left({f}_{1}âˆ—{f}_{2}\right)\left(z\right)$
(3.24)

and

$\mathrm{Ï„}=pâˆ’4p\frac{\left({A}_{1}âˆ’{B}_{1}\right)\left({A}_{2}âˆ’{B}_{2}\right)}{\left(1âˆ’{B}_{1}\right)\left(1âˆ’{B}_{2}\right)}\left[1âˆ’{\frac{1}{2}}_{2}{F}_{1}\left(1,1;\frac{pâˆ’1}{\mathrm{Î±}}+1;\frac{1}{2}\right)\right].$
(3.25)

The result is best possible when${B}_{1}={B}_{2}=âˆ’1$.

Proof Suppose that the functions ${f}_{i}âˆˆA\left(p\right)$ ($i=1,2$) satisfy the condition (3.22). Then by setting

${h}_{i}\left(z\right)=\left(1âˆ’\mathrm{Î±}\right)\frac{{\left({J}_{p,s,b}\left({f}_{i}\right)\left(z\right)\right)}^{\mathrm{â€²}}}{p{z}^{pâˆ’1}}+\mathrm{Î±}\frac{{\left({J}_{p,s,b}\left({f}_{i}\right)\left(z\right)\right)}^{\mathrm{â€²}\mathrm{â€²}}}{p\left(pâˆ’1\right){z}^{pâˆ’2}}\phantom{\rule{1em}{0ex}}\left(i=1,2\right),$
(3.26)

we have

${h}_{i}âˆˆH\left({\mathrm{Ï±}}_{i}\right),\phantom{\rule{1em}{0ex}}{\mathrm{Ï±}}_{i}=\frac{1âˆ’{A}_{\mathbf{i}}}{1âˆ’{B}_{\mathbf{i}}}\phantom{\rule{0.25em}{0ex}}\left(i=1,2\right).$

And

${\left({J}_{p,s,b}\left({f}_{\mathbf{i}}\right)\left(z\right)\right)}^{\mathrm{â€²}}=\frac{p\left(pâˆ’1\right)}{\mathrm{Î±}}{z}^{\frac{\left(1âˆ’p\right)\left(1âˆ’\mathrm{Î±}\right)}{\mathrm{Î±}}}{âˆ«}_{0}^{z}{t}^{\frac{pâˆ’1}{\mathbit{Î±}}âˆ’\mathbf{1}}{h}_{i}\left(t\right)\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{1em}{0ex}}\left(i=1,2\right),$
(3.27)

from (3.24), (3.26) and (3.27), we have

${\left({J}_{p,s,b}\left(F\right)\left(z\right)\right)}^{\mathrm{â€²}}=\frac{p\left(pâˆ’1\right)}{\mathrm{Î±}}{\mathbf{z}}^{\frac{\left(\mathbf{1}âˆ’\mathbf{p}\right)\left(\mathbf{1}âˆ’\mathbit{Î±}\right)}{\mathbit{Î±}}}{âˆ«}_{0}^{z}{t}^{\frac{pâˆ’1}{\mathbit{Î±}}âˆ’1}H\left(t\right)\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{1em}{0ex}}\left(i=1,2\right).$
(3.28)

For convenience,

$\begin{array}{rl}H\left(z\right)& =\left(1âˆ’\mathrm{Î±}\right)\frac{{\left({J}_{p,s,b}\left(F\right)\left(z\right)\right)}^{\mathrm{â€²}}}{p{z}^{pâˆ’1}}+\mathrm{Î±}\frac{{\left({J}_{p,s,b}\left(F\right)\left(z\right)\right)}^{\mathrm{â€²}\mathrm{â€²}}}{p\left(pâˆ’1\right){z}^{pâˆ’2}}\\ =\frac{p\left(pâˆ’1\right)}{\mathrm{Î±}}{z}^{\frac{\left(1âˆ’p\right)}{\mathrm{Î±}}}{âˆ«}_{0}^{z}{t}^{\frac{pâˆ’1}{\mathbit{Î±}}âˆ’1}\left({h}_{1}âˆ—{h}_{2}\right)\left(t\right)\phantom{\rule{0.2em}{0ex}}dt.\end{array}$
(3.29)

Since ${h}_{i}âˆˆH\left({\mathrm{Ï±}}_{i}\right)$ ($i=1,2$), it follows from Lemma 3 that

$\left({h}_{1}âˆ—{h}_{2}\right)\left(z\right)âˆˆH\left({\mathrm{Ï±}}_{3}\right),\phantom{\rule{1em}{0ex}}{\mathrm{Ï±}}_{3}=1âˆ’2\left(1âˆ’{\mathrm{Ï±}}_{1}\right)\left(1âˆ’{\mathrm{Ï±}}_{2}\right).$
(3.30)

By using (3.30) in (3.29) and applying Lemmas 2 and 3, we have

$\begin{array}{rcl}Re\left\{H\left(z\right)\right\}& =& \frac{p\left(pâˆ’1\right)}{\mathrm{Î±}}{âˆ«}_{0}^{1}{s}^{\frac{pâˆ’1}{\mathrm{Î±}}âˆ’1}Re\left\{\left({h}_{1}âˆ—{h}_{2}\right)\left(sz\right)\right\}\phantom{\rule{0.2em}{0ex}}ds\\ â‰¥& \frac{p\left(pâˆ’1\right)}{\mathrm{Î±}}{âˆ«}_{0}^{1}{s}^{\frac{pâˆ’1}{\mathrm{Î±}}âˆ’1}\left(2{\mathrm{Ï±}}_{3}âˆ’1+\frac{2\left(1âˆ’{\mathrm{Ï±}}_{3}\right)}{1+s|z|}\right)\phantom{\rule{0.2em}{0ex}}ds\\ >& \frac{p\left(pâˆ’1\right)}{\mathrm{Î±}}{âˆ«}_{0}^{1}{s}^{\frac{pâˆ’1}{\mathrm{Î±}}âˆ’1}\left(2{\mathrm{Ï±}}_{3}âˆ’1+\frac{2\left(1âˆ’{\mathrm{Ï±}}_{3}\right)}{1+s}\right)\phantom{\rule{0.2em}{0ex}}ds\\ =& pâˆ’4p\frac{\left({A}_{1}âˆ’{B}_{1}\right)\left({A}_{2}âˆ’{B}_{2}\right)}{\left(1âˆ’{B}_{1}\right)\left(1âˆ’{B}_{2}\right)}\left[1âˆ’\frac{pâˆ’1}{\mathrm{Î±}}{âˆ«}_{0}^{1}{s}^{\frac{pâˆ’1}{\mathrm{Î±}}âˆ’1}{\left(1+s\right)}^{âˆ’1}\phantom{\rule{0.2em}{0ex}}ds\right]\\ =& pâˆ’4p\frac{\left({A}_{1}âˆ’{B}_{1}\right)\left({A}_{2}âˆ’{B}_{2}\right)}{\left(1âˆ’{B}_{1}\right)\left(1âˆ’{B}_{2}\right)}\left[1âˆ’{\frac{1}{2}}_{2}{F}_{1}\left(1,1;\frac{pâˆ’1}{\mathrm{Î±}}+1;\frac{1}{2}\right)\right]\phantom{\rule{1em}{0ex}}\left(\mathbf{z}â†’âˆ’\mathbf{1}\right)\\ =& \mathrm{Ï„}\phantom{\rule{1em}{0ex}}\left(zâˆˆU\right).\end{array}$

When ${B}_{1}={B}_{2}=âˆ’1$, we consider ${f}_{i}âˆˆA\left(p\right)$ ($i=1,2$) satisfy the condition (3.22) and are defined by

${\left({J}_{p,s,b}\left({f}_{i}\right)\left(z\right)\right)}^{\mathrm{â€²}}=\frac{p\left(pâˆ’1\right)}{\mathrm{Î±}}{z}^{\frac{\left(1âˆ’p\right)\left(1âˆ’\mathrm{Î±}\right)}{\mathrm{Î±}}}{âˆ«}_{0}^{z}{t}^{\frac{pâˆ’1}{\mathrm{Î±}}âˆ’1}\left(\frac{1+{A}_{i}t}{1âˆ’t}\right)\phantom{\rule{0.2em}{0ex}}dt\phantom{\rule{1em}{0ex}}\left(i=1,2\right).$

By using (3.29) and applying Lemma 3, we have

$\begin{array}{rcl}H\left(z\right)& =& \frac{p\left(pâˆ’1\right)}{\mathrm{Î±}}{âˆ«}_{0}^{1}{s}^{\frac{pâˆ’1}{\mathrm{Î±}}âˆ’1}\left[1âˆ’\left(1+{A}_{1}\right)\left(1+{A}_{2}\right)+\frac{\left(1+{A}_{1}\right)\left(1+{A}_{2}\right)}{1âˆ’sz}\right]\phantom{\rule{0.2em}{0ex}}ds\\ =& pâˆ’p\left(1+{A}_{1}\right)\left(1+{A}_{2}\right)+p\left(1+{A}_{1}\right)\left(1+{A}_{2}\right)\left(1âˆ’z\right)^{âˆ’1}{}_{2}{F}_{1}\left(1,1;\frac{pâˆ’1}{\mathrm{Î±}}+1;\frac{z}{zâˆ’1}\right)\\ â†’& pâˆ’p\left(1+{A}_{1}\right)\left(1+{A}_{2}\right)+\frac{p}{2}\left(1+{A}_{1}\right){\left(1+{A}_{2}\right)}_{2}{F}_{1}\left(1,1;\frac{pâˆ’1}{\mathrm{Î±}}+1;\frac{1}{2}\right)\phantom{\rule{1em}{0ex}}\left(zâ†’âˆ’1\right).\end{array}$

This completes the proof of Theorem 5.â€ƒâ–¡

Remark 1 Putting ${A}_{i}=1âˆ’2{\mathrm{Î¸}}_{i}$ ($0â‰¤{\mathrm{Î¸}}_{i}<1$) and ${B}_{i}=âˆ’1$ ($i=1,2$) in Theorem 5, we obtain the result obtained by Liu [10], Theorem 5].

Putting ${A}_{i}=1âˆ’2{\mathrm{Î¸}}_{i}$ ($0â‰¤{\mathrm{Î¸}}_{i}<1$), ${B}_{i}=âˆ’1$ ($i=1,2$) and $s=0$ in Theorem 5, we obtain the following corollary.

Corollary 2 Let$\mathrm{Ï‡}and${f}_{i}âˆˆA\left(p\right)$satisfy the following inequality:

$Re\left\{\left(1âˆ’\mathrm{Î±}\right)\frac{{f}_{i}^{\mathrm{â€²}}\left(z\right)}{p{z}^{pâˆ’1}}+\mathrm{Î±}\frac{{f}_{i}^{\mathrm{â€²}\mathrm{â€²}}\left(z\right)}{p\left(pâˆ’1\right){z}^{pâˆ’2}}\right\}>{\mathrm{Î¸}}_{i}\phantom{\rule{1em}{0ex}}\left(0â‰¤{\mathrm{Î¸}}_{i}<1;i=1,2\right),$

then

$Re\left\{\left(1âˆ’\mathrm{Î±}\right)\frac{{\left({f}_{1}âˆ—{f}_{2}\right)}^{\mathrm{â€²}}\left(z\right)}{p{z}^{pâˆ’1}}+\mathrm{Î±}\frac{{\left({f}_{1}âˆ—{f}_{2}\right)}^{\mathrm{â€²}\mathrm{â€²}}\left(z\right)}{p\left(pâˆ’1\right){z}^{pâˆ’2}}\right\}>\frac{\mathrm{Ï‡}}{p},$

where

$\mathrm{Ï‡}=pâˆ’4p\left(1âˆ’{\mathrm{Î¸}}_{1}\right)\left(1âˆ’{\mathrm{Î¸}}_{2}\right)\left[1âˆ’{\frac{1}{2}}_{2}{F}_{1}\left(1,1;\frac{pâˆ’1}{\mathrm{Î±}}+1;\frac{1}{2}\right)\right].$

The result is best possible.

Theorem 6 Let$fâˆˆ{S}_{p}^{s,b}\left(A,B\right)$and$gâˆˆA\left(p\right)$satisfy the following inequality:

$Re\left\{\frac{g\left(z\right)}{{z}^{p}}\right\}>\frac{1}{2}\phantom{\rule{1em}{0ex}}\left(zâˆˆU\right),$
(3.31)

then

$\left(fâˆ—g\right)\left(z\right)âˆˆ{S}_{p}^{s,b}\left(A,B\right).$

Proof We have

$\frac{{\left({J}_{p,s,b}\left(fâˆ—g\right)\left(z\right)\right)}^{\mathrm{â€²}}}{p{z}^{pâˆ’1}}=\frac{{\left({J}_{p,s,b}\left(f\right)\left(z\right)\right)}^{\mathrm{â€²}}}{p{z}^{pâˆ’1}}âˆ—\frac{g\left(z\right)}{{z}^{p}}\phantom{\rule{1em}{0ex}}\left(zâˆˆU\right),$

where $g\left(z\right)$ satisfies (3.31) and $\frac{1+Az}{1+Bz}$ is convex (univalent) in U. By using (1.10) and applying Lemma 5, we complete the proof of Theorem 6.â€ƒâ–¡

Theorem 7 Let$\mathrm{Ïƒ}>0$and$fâˆˆA\left(p\right)$satisfy the following subordination condition:

$\left(1âˆ’\mathrm{Î±}\right)\frac{{J}_{p,s,b}\left(f\right)\left(z\right)}{{z}^{p}}+\mathrm{Î±}\frac{{\left({J}_{p,s,b}\left(f\right)\left(z\right)\right)}^{\mathrm{â€²}}}{p{z}^{pâˆ’1}}â‰º\frac{1+Az}{1+Bz}.$
(3.32)

Then

$Re{\left\{\frac{{J}_{p,s,b}\left(f\right)\left(z\right)}{{z}^{p}}\right\}}^{\frac{1}{\mathrm{Ïƒ}}}>{\mathrm{Î³}}^{\frac{1}{\mathrm{Ïƒ}}},$

where

The result is best possible.

Proof Let

$M\left(z\right)=\frac{{J}_{p,s,b}\left(f\right)\left(z\right)}{{z}^{p}}\phantom{\rule{1em}{0ex}}\left(zâˆˆU\right),$
(3.33)

where M is of the form (2.1) and is analytic in U. Differentiating (3.33) with respect to z, we have

$\left(1âˆ’\mathrm{Î±}\right)\frac{{J}_{p,s,b}\left(f\right)\left(z\right)}{{z}^{p}}+\mathrm{Î±}\frac{{\left({J}_{p,s,b}\left(f\right)\left(z\right)\right)}^{\mathrm{â€²}}}{p{z}^{pâˆ’1}}=M\left(z\right)+\frac{\mathrm{Î±}}{p}z{M}^{\mathrm{â€²}}\left(z\right)â‰º\frac{1+Az}{1+Bz}.$

Now, by following steps similar to the proof of Theorem 1 and using the elementary inequality

$Re\left\{{\mathrm{Ï’}}^{1/\mathrm{Ï°}}\right\}â‰¥{\left\{Re\mathrm{Ï’}\right\}}^{1/\mathrm{Ï°}}\phantom{\rule{1em}{0ex}}\left(Re\left\{\mathrm{Ï’}\right\}>0;\mathrm{Ï°}âˆˆ\mathbb{N}\right),$

we obtain the result asserted by Theorem 7.â€ƒâ–¡

Theorem 8 Let$\mathrm{Î½}âˆˆ{\mathbb{C}}^{âˆ—}$and$A,Bâˆˆ\mathbb{C}$withand$|B|â‰¤1$. Suppose that

If$fâˆˆA\left(p\right)$withfor all$zâˆˆ{U}^{âˆ—}=U\mathrm{â•²}\left\{0\right\}$, then

$\frac{{J}_{p,sâˆ’1,b}\left(f\right)\left(z\right)}{{J}_{p,s,b}\left(f\right)\left(z\right)}â‰º\frac{1+Az}{1+Bz},$

implies

${\left(\frac{{J}_{p,s,b}\left(f\right)\left(z\right)}{{z}^{p}}\right)}^{\mathrm{Î½}}â‰º{q}_{1}\left(z\right),$

where

is the best dominant.

Proof Let us put

$\mathrm{Ï†}\left(z\right)={\left(\frac{{J}_{p,s,b}\left(f\right)\left(z\right)}{{z}^{p}}\right)}^{\mathrm{Î½}}\phantom{\rule{1em}{0ex}}\left(zâˆˆU\right).$
(3.34)

Then Ï† is analytic in U, $\mathrm{Ï†}\left(0\right)=1$ and for all $zâˆˆU$. Taking the logarithmic derivatives in both sides of (3.34) and using the identity (1.9), we have

$1+\frac{z{\mathrm{Ï†}}^{\mathrm{â€²}}\left(z\right)}{\mathrm{Î½}\left(b+1\right)\mathrm{Ï†}\left(z\right)}=\frac{{J}_{p,sâˆ’1,b}\left(f\right)\left(z\right)}{{J}_{p,s,b}\left(f\right)\left(z\right)}â‰º\frac{1+Az}{1+Bz}.$

Now the assertions of Theorem 8 follow by using Lemma 6 for $\mathrm{Î³}=\mathrm{Î½}\left(b+1\right)$.â€ƒâ–¡

Putting $B=âˆ’1$ and $A=1âˆ’2\mathrm{Ïƒ}$, $0â‰¤\mathrm{Ïƒ}<1$, in Theorem 8, we obtain the following corollary.

Corollary 3 Assume that$\mathrm{Î½}âˆˆ{\mathbb{C}}^{âˆ—}$satisfies either$|2\mathrm{Î½}\left(b+1\right)\left(1âˆ’\mathrm{Ïƒ}\right)âˆ’1|â‰¤1$or$|2\mathrm{Î½}\left(b+1\right)\left(1âˆ’\mathrm{Ïƒ}\right)+1|â‰¤1$. If$fâˆˆA\left(p\right)$withfor$zâˆˆ{U}^{âˆ—}$, then

$Re\left\{\frac{{J}_{p,sâˆ’1,b}\left(f\right)\left(z\right)}{{J}_{p,s,b}\left(f\right)\left(z\right)}\right\}>\mathrm{Ïƒ}\phantom{\rule{1em}{0ex}}\left(zâˆˆU\right),$

implies

${\left(\frac{{J}_{p,s,b}\left(f\right)\left(z\right)}{{z}^{p}}\right)}^{\mathrm{Î½}}â‰º{q}_{2}\left(z\right)={\left(1âˆ’z\right)}^{âˆ’2\mathrm{Î½}\left(b+1\right)\left(1âˆ’\mathrm{Ïƒ}\right)},$

and${q}_{2}$is the best dominant.

Remark 2 Specializing the parameters s and b in the above results of this paper, we obtain the results for the corresponding operators ${F}_{\mathrm{Î¼},p}$, ${I}_{p}^{\mathrm{Î±}}$, ${J}_{p}^{\mathrm{Î³}}$ and ${J}_{p}^{\mathrm{Î³}}\left(l\right)$ which are defined in the introduction.

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Aouf, M., Mostafa, A., Shahin, A. et al. Applications of differential subordinations for certain classes of p-valent functions associated with generalized Srivastava-Attiya operator. J Inequal Appl 2012, 153 (2012). https://doi.org/10.1186/1029-242X-2012-153

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