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Article

# Geometric Properties of Certain Analytic Functions Associated with the Dziok-Srivastava Operator

by
Cai-Mei Yan
1 and
Jin-Lin Liu
2,*
1
Information Engineering College, Yangzhou University, Yangzhou 225002, China
2
Department of Mathematics, Yangzhou University, Yangzhou 225002, China
*
Author to whom correspondence should be addressed.
Symmetry 2019, 11(2), 259; https://doi.org/10.3390/sym11020259
Submission received: 20 January 2019 / Revised: 13 February 2019 / Accepted: 15 February 2019 / Published: 19 February 2019
(This article belongs to the Special Issue Integral Transforms and Operational Calculus)

## Abstract

:
The objective of the present paper is to derive certain geometric properties of analytic functions associated with the Dziok–Srivastava operator.
2010 Mathematics Subject Classification:
30C45

## 1. Introduction

Throughout this paper, we assume that:
$n , p ∈ N , − 1 ≤ B < A ≤ 1 , α > 0 and β < 1 .$
Let $A n ( p )$ denote the class of functions of the form:
$f ( z ) = z p + ∑ k = n + p ∞ a k z k$
which are analytic in the open unit disk $U = { z : | z | < 1 }$. If $f ( z ) = z p + ∑ k = n + p ∞ a k z k ∈ A n ( p )$ and $g ( z ) = z p + ∑ k = n + p ∞ b k z k ∈ A n ( p )$, then the Hadamard product (or convolution) of f and g is defined by:
$( f * g ) ( z ) = z p + ∑ k = n + p ∞ a k b k z k$
For:
$α j ∈ C ( j = 1 , 2 , ⋯ , l ) and β j ∈ C \ { 0 , − 1 , − 2 , ⋯ } ( j = 1 , 2 , ⋯ , m )$
the generalized hypergeometric function $l F m ( α 1 , ⋯ , α l ; β 1 , ⋯ , β m ; z )$ is defined by:
$l F m ( α 1 , ⋯ , α l ; β 1 , ⋯ , β m ; z ) = ∑ k = 0 ∞ ( α 1 ) k ⋯ ( α l ) k ( β 1 ) k ⋯ ( β m ) k z k k !$
$( l ≤ m + 1 ; l , m ∈ N 0 = N ∪ { 0 } ; z ∈ U )$
where $( x ) k$ is the Pochhammer symbol given by $( x ) k = x ( x + 1 ) ⋯ ( x + k − 1 )$ for $k ∈ N$ and $( x ) 0 = 1$. Corresponding to the function $z p l F m ( α 1 , ⋯ , α l ; β 1 , ⋯ , β m ; z )$, the well-known Dziok–Srivastava operator [1] $H ( α 1 , ⋯ , α l ; β 1 , ⋯ , β m ) : A n ( p ) → A n ( p )$ is defined by:
$H ( α 1 , ⋯ , α l ; β 1 , ⋯ , β m ) f ( z ) = z p l F m ( α 1 , ⋯ , α l ; β 1 , ⋯ , β m ; z * f ( z )$
$( l ≤ m + 1 ; l , m ∈ N 0 ; z ∈ U )$
If $f ∈ A n ( p )$ is given by (2), then we have:
$H ( α 1 , ⋯ , α l ; β 1 , ⋯ , β m ) f ( z ) = z p + ∑ k = n + p ∞ ( α 1 ) k ⋯ ( α l ) k ( β 1 ) k ⋯ ( β m ) k a k k ! z k$
For convenience, we write:
$H m l ( α 1 ) = H ( α 1 , ⋯ , α l ; β 1 , ⋯ , β m ) ( l ≤ m + 1 ; l , m ∈ N 0 )$
It is noteworthy to mention that the Dziok–Srivastava operator is a generalization of certain linear operators considered in earlier investigations.
Next, we consider the function $h ( A , B ; z ) = ( 1 + A z ) / ( 1 + B z )$ for $z ∈ U$. It is known that the function $h ( A , B ; z )$ is the conformal map of U onto a disk, symmetrical with respect to the real axis, which is centered at the point $( 1 − A B ) / ( 1 − B 2 )$$( B ≠ ± 1 )$ and with its radius equal to $( A − B ) /$$( 1 − B 2 )$$( B ≠ ± 1 )$. Furthermore, the boundary circle of this disk intersects the real axis at the points $( 1 − A ) / ( 1 − B )$ and $( 1 + A ) / ( 1 + B )$ with $B ≠ ± 1$.
Let $P [ A , B ]$ denote the class of functions of the form $p ( z ) = 1 + p 1 z + ⋯$, which are analytic in U and satisfy the subordination $p ( z ) ≺ h ( A , B ; z )$. It is clear that $p ∈ P [ A , B ]$ if and only if:
$p ( z ) − 1 − A B 1 − B 2 < A − B 1 − B 2 ( − 1 < B < A ≤ 1 ; z ∈ U )$
and:
$Re p ( z ) > 1 − A 2 ( B = − 1 ; z ∈ U )$
For two functions f and g analytic in U, f is said to be subordinate to g, written by $f ( z ) ≺ g ( z ) ( z ∈ U )$, if there exists a Schwarz function w in U such that:
$| w ( z ) | ≤ | z | and f ( z ) = g ( w ( z ) ) ( z ∈ U )$
Furthermore, if the function g is univalent in U, then:
$f ( z ) ≺ g ( z ) ( z ∈ U ) ⟺ f ( 0 ) = g ( 0 ) and f ( U ) ⊂ g ( U )$
Many properties of analytic functions have been investigated by several authors(see [1,2,3,4,5,6,7,8,9,10,11]). In this paper, we derive certain geometric properties of analytic functions associated with the well-known Dziok–Srivastava operator.

## 2. Main Results

Theorem 1.
Let f belong to the class $A n ( p )$. Furthermore, let:
$H m l ( α 1 ) f ( z ) z p ∈ P [ A , B ] .$
Then:
where:
$K A = 1 − A 2 r 2 n + n A r n − 1 ( 1 − r 2 ) , K B = 1 − B 2 r 2 n + n B r n − 1 ( 1 − r 2 ) , L n = 2 α ( 1 − A B r 2 n ) + n α ( A + B ) r n − 1 ( 1 − r 2 ) + ( A − B ) r n − 1 ( 1 − r 2 ) , M n ( A , B , α , r ) = 2 α K B ( 1 + A r n ) − L n ( 1 + B r n ) .$
The result is sharp.
Proof.
For $z = 0$, the equality in (4) holds true. Thus, we assume that $0 < | z | = r < 1$. From (3), we can write:
$H m l ( α 1 ) f ( z ) z p = 1 + A z n φ ( z ) 1 + B z n φ ( z ) ( z ∈ U ) ,$
where $φ ( z )$ is analytic and $| φ ( z ) | ≤ 1$ in U. From (7), we have:
$H m l ( α 1 ) f ( z ) z p + α z ( H m l ( α 1 ) f ( z ) z p ) ′ = H m l ( α 1 ) f ( z ) z p + α ( A − B ) z n ( n φ ( z ) + z φ ′ ( z ) ) ( 1 + B z n φ ( z ) ) 2 = H m l ( α 1 ) f ( z ) z p + n α A − B ( A − B H m l ( α 1 ) f ( z ) / z p ) ( H m l ( α 1 ) f ( z ) / z p − 1 ) + α ( A − B ) z n + 1 φ ′ ( z ) ( 1 + B z n φ ( z ) ) 2$
By using the Carathéodory inequality:
$| φ ′ ( z ) | ≤ 1 − | φ ( z ) | 2 1 − r 2 ,$
we get:
$Re z n + 1 φ ′ ( z ) ( 1 + B z n φ ( z ) ) 2 ≤ r n + 1 ( 1 − | φ ( z ) | 2 ) ( 1 − r 2 ) | 1 + B z n φ ( z ) | 2 = r 2 n | A − B H m l ( α 1 ) f ( z ) / z p | 2 − | H m l ( α 1 ) f ( z ) / z p − 1 | 2 ( A − B ) 2 r n − 1 ( 1 − r 2 )$
Set $H m l ( α 1 ) f ( z ) z p = u + i v ( u , v ∈ R )$. Then, (8) and (9) give:
$Re H m l ( α 1 ) f ( z ) z p + α z ( H m l ( α 1 ) f ( z ) z p ) ′ ≤ 1 + n α A + B A − B u − n α A A − B − n α B A − B ( u 2 − v 2 ) + α r 2 n ( ( A − B u ) 2 + ( B v ) 2 ) − ( ( u − 1 ) 2 + v 2 ) ( A − B ) r n − 1 ( 1 − r 2 ) = 1 + n α A + B A − B u − n α A − B ( A + B u 2 ) + α r 2 n ( A − B u ) 2 − ( u − 1 ) 2 ( A − B ) r n − 1 ( 1 − r 2 ) + α A − B n B − 1 − B 2 r 2 n r n − 1 ( 1 − r 2 ) v 2$
Note that:
$1 − B 2 r 2 n r n − 1 ( 1 − r 2 ) ≥ 1 − r 2 n r n − 1 ( 1 − r 2 ) = 1 r n − 1 ( 1 + r 2 + r 4 + ⋯ + r 2 ( n − 2 ) + r 2 ( n − 1 ) ) = 1 2 r n − 1 [ ( 1 + r 2 ( n − 1 ) ) + ( r 2 + r 2 ( n − 2 ) ) + ⋯ + ( r 2 ( n − 1 ) + 1 ) ] ≥ n ≥ n B$
Using (10) and (11), we obtain:
$Re { H m l ( α 1 ) f ( z ) z p + α z ( H m l ( α 1 ) f ( z ) z p ) ′ } ≤ 1 + n α A + B A − B u − n α A − B ( A + B u 2 ) + α r 2 n ( A − B u ) 2 − ( u − 1 ) 2 ( A − B ) r n − 1 ( 1 − r 2 ) = ψ n ( u )$
It is known that for $| ξ | ≤ σ ( σ < 1 )$,
$1 + A ξ 1 + B ξ − 1 − A B σ 2 1 − B 2 σ 2 ≤ ( A − B ) σ 1 − B 2 σ 2$
and:
$1 − A σ 1 − B σ ≤ Re 1 + A ξ 1 + B ξ ≤ 1 + A σ 1 + B σ$
Furthermore, (7) and (14) show that:
$1 − A r n 1 − B r n ≤ Re { H m l ( α 1 ) f ( z ) z p } ≤ 1 + A r n 1 + B r n .$
Now, we calculate the maximum value of $ψ n ( u )$ on the segment $1 − A r n 1 − B r n , 1 + A r n 1 + B r n$. Obviously,
$ψ n ′ ( u ) = 1 + n α A + B A − B − 2 n α B A − B u + 2 α ( 1 − A B r 2 n ) − ( 1 − B 2 r 2 n ) u ( A − B ) r n − 1 ( 1 − r 2 )$
and $ψ n ′ ( u ) = 0$ if and only if:
Since:
$2 α K B ( 1 − A r n ) − L n ( 1 − B r n ) = 2 α [ ( 1 − A r n ) ( 1 − B 2 r 2 n ) − ( 1 − B r n ) ( 1 − A B r 2 n ) ] − n α r n − 1 ( 1 − r 2 ) [ ( A + B ) ( 1 − B r n ) − 2 B ( 1 − A r n ) ] − ( A − B ) r n − 1 ( 1 − r 2 ) ( 1 − B r n ) = − 2 α ( A − B ) r n ( 1 − B r n ) − n α ( A − B ) r n − 1 ( 1 − r 2 ) ( 1 + B r n ) − ( A − B ) r n − 1 ( 1 − r 2 ) ( 1 − B r n ) < 0$
we see that:
$u n > 1 − A r n 1 − B r n$
However, $u n$ is not always less than $1 + A r n 1 + B r n$. The following two cases arise.
Case (I). $u n ≥ 1 + A r n 1 + B r n$, that is $M n ( A , B , α , r )$$≤ 0$. In view of $ψ n ′ ( u n ) = 0$ and (15), the function $ψ n ( u )$ is increasing on the segment $1 − A r n 1 − B r n , 1 + A r n 1 + B r n$. Thus, we deduce from (12) that, if $M n ( A , B , α , r ) ≤ 0$, then:
$Re H m l ( α 1 ) f ( z ) z p + α z ( H m l ( α 1 ) f ( z ) z p ) ′ ≤ ψ n 1 + A r n 1 + B r n = 1 + n α A + B A − B 1 + A r n 1 + B r n − n α A − B A + B 1 + A r n 1 + B r n 2 = 1 + A r n 1 + B r n − n α A − B 1 − 1 + A r n 1 + B r n A − B 1 + A r n 1 + B r n = 1 + ( A + B + n α ( A − B ) ) r n + A B r 2 n ( 1 + B r n ) 2$
This gives (4).
Next, we consider the function f defined by:
$H m l ( α 1 ) f ( z ) z p = 1 + A z n 1 + B z n$
which satisfies the condition (3). It is easy to check that:
$H m l ( α 1 ) f ( r ) r p + α r ( H m l ( α 1 ) f ( r ) r p ) ′ = 1 + ( A + B + n α ( A − B ) ) r n + A B r 2 n ( 1 + B r n ) 2$
which implies that the inequality (4) is sharp.
Case (II). $u n ≤ 1 + A r n 1 + B r n$, that is $M n ( A , B , α , r ) ≥ 0$. In this case, we easily have:
$Re H m l ( α 1 ) f ( z ) z p + α z ( H m l ( α 1 ) f ( z ) z p ) ′ ≤ ψ n ( u n )$
In view of (6), $ψ n ( u )$ in (12) can be written as:
$ψ n ( u ) = − α K B u 2 + L n u − α K A ( A − B ) r n − 1 ( 1 − r 2 )$
Therefore, if $M n ( A , B , α , r ) ≥ 0$, then it follows from (16), (18), and (19) that:
$Re H m l ( α 1 ) f ( z ) z p + α z ( H m l ( α 1 ) f ( z ) z p ) ′ ≤ − α K B u n 2 + L n u n − α K A ( A − B ) r n − 1 ( 1 − r 2 ) = L n 2 − 4 α 2 K A K B 4 α ( A − B ) r n − 1 ( 1 − r 2 ) K B$
To show the sharpness, we take:
$H m l ( α 1 ) f ( z ) z p = 1 + A z n φ ( z ) 1 + B z n φ ( z ) and φ ( z ) = z − c n 1 − c n z$
where $c n ∈ R$ is determined by:
$H m l ( α 1 ) f ( r ) r p = 1 + A r n φ ( r ) 1 + B r n φ ( r ) = u n ∈ 1 − A r n 1 − B r n , 1 + A r n 1 + B r n$
Clearly, $− 1 < φ ( r ) ≤ 1 , − 1 ≤ c n < 1 , | φ ( z ) | ≤ 1 ( z ∈ U )$, and so, f satisfies the condition (3). Since:
$φ ′ ( r ) = 1 − c n 2 ( 1 − c n r ) 2 = 1 − | φ ( r ) | 2 1 − r 2$
from the above argument, we find that:
$H m l ( α 1 ) f ( r ) r p + α r ( H m l ( α 1 ) f ( r ) z p ) ′ = ψ n ( u n )$
The proof of the theorem is now completed. □
Corollary 1.
Let $f ∈ A 1 ( p )$, and satisfy $Re { H m l ( α 1 ) f ( z ) / z p } > β$$( β < 1 ; z ∈ U )$. Then, for $| z | = r < 1$,
$Re H m l ( α 1 ) f ( z ) z p + α z ( H m l ( α 1 ) f ( z ) z p ) ′ ≤ β + ( 1 − β ) 1 + 2 α r − r 2 ( 1 − r ) 2$
The result is sharp.
Proof.
By considering $H m l ( α 1 ) f ( z ) / z p − β 1 − β$ instead of $H m l ( α 1 ) f ( z ) / z p$, we only need to prove the corollary for $β = 0$. Putting $n = A = 1$ and $B = − 1$ in (6), we get:
$K 1 = 2 ( 1 − r 2 ) , K − 1 = 0 , L 1 = 2 α ( 1 + r 2 ) + 2 ( 1 − r 2 )$
and:
$M 1 ( 1 , − 1 , α , r ) = − 2 ( 1 − r ) [ 1 + α − ( 1 − α ) r 2 ] ≤ 0$
Consequently, an application of (4) in Theorem 2.1 yields:
$Re H m l ( α 1 ) f ( z ) z p + α z ( H m l ( α 1 ) f ( z ) z p ) ′ ≤ 1 + 2 α r − r 2 ( 1 − r ) 2$
The sharpness follows immediately from that of Theorem 1. □
Theorem 2.
Let $α j$$( j = 1 , 2 , ⋯ , l )$ and $β s$$( s = 1 , 2 , ⋯ , m )$ be positive real numbers. Furthermore, let $f ( z ) = z p + ∑ k = n + p ∞ a k z k ∈ A n ( p )$, and satisfy:
$H m l ( α 1 ) f ( z ) z p + α z H m l ( α 1 ) f ( z ) z p ′ ∈ P [ A , B ]$
Then:
$| a k | ≤ k ! ( A − B ) ( β 1 ) k ⋯ ( β m ) k ( 1 + α ( k − p ) ) ( α 1 ) k ⋯ ( α l ) k ( k ≥ n + p )$
The result is sharp for each $k ≥ n + p$.
Proof.
It is well known that if:
$g ( z ) = ∑ k = 1 ∞ b k z k ≺ φ ( z ) ( z ∈ U )$
where $g ( z )$ is analytic in U and $φ ( z ) = z + ⋯$ is convex univalent in U, then $| b k | ≤ 1$$( k = 1 , 2 , 3 , ⋯ )$.
From (20), we have:
$1 A − B H m l ( α 1 ) f ( z ) z p + α z H m l ( α 1 ) f ( z ) z p ′ − 1 = 1 A − B ∑ k = n + p ∞ ( 1 + α ( k − p ) ) ( α 1 ) k ⋯ ( α l ) k · a k k ! ( β 1 ) k ⋯ ( β m ) k z k − p ≺ z 1 + B z ( z ∈ U )$
In view of the function $z 1 + B z$ being convex univalent in U, it follows from (22) that:
$( 1 + α ( k − p ) ) ( α 1 ) k ⋯ ( α l ) k k ! ( A − B ) ( β 1 ) k ⋯ ( β m ) k | a k | ≤ 1 ( k ≥ n + p )$
which gives (21).
Next, we consider the function $f k − p ( z )$ defined by:
$f k − p ( z ) = z p + ( A − B ) ∑ q = 1 ∞ ( − B ) q − 1 ( β 1 ) q k ⋯ ( β m ) q k ( q k ) ! ( 1 + α q ( k − p ) ) ( α 1 ) q k ⋯ ( α l ) q k z q ( k − p ) + p ( z ∈ U ; k ≥ n + p )$
Since:
$H m l ( α 1 ) f k − p ( z ) z p + α z H m l ( α 1 ) f k − p ( z ) z p ′ = 1 + A z k − p 1 + B z k − p ≺ 1 + A z 1 + B z ( z ∈ U )$
and:
$f k − p ( z ) = z p + k ! ( A − B ) ( β 1 ) k ⋯ ( β m ) k ( 1 + α ( k − p ) ) ( α 1 ) k ⋯ ( α l ) k z k + ⋯$
for each $k ≥ n + p$, the proof of Theorem 2 is completed. □

## Author Contributions

All authors contributed equally.

## Funding

This work is supported by the National Natural Science Foundation of China (Grant No. 11571299).

## Conflicts of Interest

The authors declare no conflict of interest.

## References

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MDPI and ACS Style

Yan, C.-M.; Liu, J.-L. Geometric Properties of Certain Analytic Functions Associated with the Dziok-Srivastava Operator. Symmetry 2019, 11, 259. https://doi.org/10.3390/sym11020259

AMA Style

Yan C-M, Liu J-L. Geometric Properties of Certain Analytic Functions Associated with the Dziok-Srivastava Operator. Symmetry. 2019; 11(2):259. https://doi.org/10.3390/sym11020259

Chicago/Turabian Style

Yan, Cai-Mei, and Jin-Lin Liu. 2019. "Geometric Properties of Certain Analytic Functions Associated with the Dziok-Srivastava Operator" Symmetry 11, no. 2: 259. https://doi.org/10.3390/sym11020259

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