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Article

# Hadamard Product of Certain Multivalent Analytic Functions with Positive Real Parts

1
Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia
2
Department of Mathematics, Faculty of Science, Mansoura University, Mansoura 35516, Egypt
*
Author to whom correspondence should be addressed.
Mathematics 2022, 10(9), 1506; https://doi.org/10.3390/math10091506
Received: 2 April 2022 / Revised: 28 April 2022 / Accepted: 29 April 2022 / Published: 1 May 2022
(This article belongs to the Special Issue Complex Analysis and Geometric Function Theory)

## Abstract

:
This paper aims to provide sufficient conditions for starlikeness and convexity of Hadamard product (convolution) of certain multivalent analytic functions with positive real parts. Moreover, the starlikeness conditions for a certain integral operator and other convolution results are also considered.
MSC:
30C45; 30C50; 30C80

## 1. Introduction

Let $A p$ denote the class of functions of the form:
$f ( z ) = z p + ∑ k = 1 ∞ a k + p z k + p ( p ∈ N = { 1 , 2 , … } )$
which are analytic in the open unit disc $U = { z : z < 1 }$ and let $A 1 : = A .$ A function $f ∈ A p$ is said to be in the class $S p *$ of p-valently starlike functions in $U$ if it satisfies the following inequality:
$ℜ ( z f ′ ( z ) f ( z ) ) > 0 ( z ∈ U ) .$
Further, A function $f ∈ A p$ is said to be in the class $K p$ of p-valently convex in $U$ if it satisfies the following inequality:
$ℜ ( 1 + z f ″ ( z ) f ′ ( z ) ) > 0 ( z ∈ U ) .$
The starlikeness and convexity of $p$-valent functions were introduced by Goodman [1] and considered recently in the works [2,3,4,5,6,7,8,9,10,11,12]. Let $P α$ be the class of functions with positive real part of order $α$ that have the form $h ( z ) = 1 + ∑ k = 1 ∞ c k z k$ which are analytic in $U$ and satisfy the following condition
$ℜ h ( z ) > α , ( 0 ≤ α < 1 ; z ∈ U ) .$
A function $f ∈ A p$ is said to be in the class $P ( p , α )$ if and only if
$f ′ ( z ) p z p − 1 ∈ P α ( 0 ≤ α < 1 ; z ∈ U ) .$
For $0 ≤ α < 1 ,$ we denote by $R p ( α )$ the family of functions $f ∈ A p$ which satisfy the condition
$f ′ ( z ) + z f ″ ( z ) p 2 z p − 1 ∈ P α ( z ∈ U ) .$
As a special case, for $p = 1$ the class $R p ( α )$ reduces to the familiar class R which was studied by Chichra [13], Ali and Thomas [14], Singh and Singh [15,16], Kim and Srivastava [17], Ali et al. [18], Szasz [19] and Yang and Liu [20]. For two functions f and $g ∈ A p$, that is if f is given by (1) and g is given by $f ( z ) = z p + ∑ k = 1 ∞ b k + p z k + p$, then their Hadamard product (convolution), $( f ∗ g )$, is the function defined by the power series
$( f ∗ g ) ( z ) = z p + ∑ k = 1 ∞ a k + p b k + p z k + p .$
For a function $f ∈ A p$, Reddy and Padmanabhan [21] defined the following integral operator:
$J p , c ( z ) = J p , c ( f ( z ) ) = c + p z c ∫ 0 z t c − 1 f ( t ) d t ( p ∈ N , c > − p ) = z p + ∑ k = 1 ∞ c + p c + p + k a k + p z k + p .$
In particular, The operator $J 1 , c$ was introduced by Bernardi [22] and the operator $J 1 , 1$ was studied earlier by Libera [23]. By using the Clunie-Jack Lemma [24] it was shown in [25] that if the function $f ∈ A$ belongs to the class $P β ,$ then $J 1 , c ∈ S *$ ($S *$ is the class of starllike functions) provided
$( 1 + c ) β > log 4 e 6 ( c 2 tan 2 α * π 2 − 3 )$
where $1 =$$α * + 2 π tan − 1 α *$. In their paper [14], Ali and Thomas improved the constant $β$ in (5). In the work of Lashin [26] a criterion for convolution properties of functions of the class $P ( α )$ was introduced, this criterion was improved by Sokol [27] and Ponnusamy and Singh [28]. The present paper extends and improves each of these earlier results in [26,27,28]. Additionally, By using Miller and Mocanu Theorem [29] we will consider the starlikeness of the integral operator $J p , c$ and extend the results of Ali and Thomas [14].

## 2. Preliminaries Lemmas

In this paper, we shall require the following lemmas.
Lemma 1
(see [15]). A sequence $b k k = 0 ∞$ of non-negative numbers is said to be a convex null sequence if $b k → 0$ as $k → ∞$ and
$b 0 − b 1 ≥ b 1 − b 2 ≥ . . . ≥ b k − b k + 1 ≥ 0 .$
Let the sequence $b k k = 0 ∞$ be a convex null sequence. Then the function
$q ( z ) = b 0 2 + ∑ k = 1 ∞ b k z k ( z ∈ U )$
is analytic in $U$ and $ℜ q ( z ) > 0 .$
Lemma 2
([15]). If the function $χ ( z )$ is analytic in $U$ with $χ ( 0 ) = 1$ and $ℜ χ ( z ) > 1 / 2 , z ∈ U ,$ then for any function F analytic in $U$, the function $χ ∗ F$ takes its values in the convex hull of $F ( U )$.
Lemma 3
([25,30]). Let $λ > 0$ and $0 ≤ β < 1 .$ If the function q is analytic in $U$ with $q ( 0 ) = 1 ,$ satisfies the inequality
$ℜ q ( z ) + λ z q ′ ( z ) > β ( z ∈ U ) ,$
then
$ℜ q ( z ) > 1 + 2 ( 1 − β ) ∑ k = 1 ∞ ( − 1 ) k 1 + λ k ( z ∈ U ) .$
Lemma 4
([31]). For $0 ≤ α < 1$ and $0 ≤ β < 1 ,$
$P α ∗ P β ⊂ P δ , δ = 1 − 2 ( 1 − α ) ( 1 − β ) .$
The result is sharp.
Lemma 5
([29]). Suppose that the function $φ : C 2 × U → C$ satisfies the condition $ℜ { φ ( i x , y ; z ) } ≤ δ$ for all real $x , y ≤ − ( 1 + x 2 ) 2$ and all $z ∈ U .$ If $q ( z ) = 1 + c 1 z + ⋯$ is analytic in U and
$ℜ { φ ( q ( z ) , z q ′ ( z ) , z ) } > δ , for z ∈ U ,$
then $ℜ { q ( z ) } > 0$ in $U .$
Lemma 6
([32]). The nth partial sum $S n$ of the Alternating series $∑ k = 1 ∞ ( − 1 ) n a n , a n > 0 ,$ always lies between $S n − 1$ and $S n − 2 ,$ or
$− 1 < − a 1 < S n < a 2 − a 1 < 0 .$

## 3. Main Results

First of all, we state and prove the following results which extend the results of Lashin [26] and Sokol [27].
Theorem 1.
Let $p ∈ N , 0 ≤ α , β < 1 ,$ and let $ψ ( p ) = ∑ k = 1 ∞ ( − 1 ) k p + k .$ If $f , g ∈ A p$ satisfy f $∈ P ( p , α )$ and $g ∈ P ( p , β ) ,$ then $ξ = ( f ∗ g ) ∈ S p * ,$ provided that
$( 1 − α ) ( 1 − β ) < min { 2 p + 1 8 p 2 ψ 2 + 4 p , p + 1 4 p 2 ( 1 − ln 4 e ) } .$
Proof.
It is easy to see that,
$f ′ ( z ) p z p − 1 ∗ g ′ ( z ) p z p − 1 = ξ ′ ( z ) + z ξ ″ ( z ) p 2 z p − 1 .$
By the hypothesis on f and $g ,$ it follows from (8) and Lemma 4 that
$ℜ ξ ′ ( z ) + z ξ ″ ( z ) p 2 z p − 1 > 1 − 2 ( 1 − α ) ( 1 − β ) .$
Let
$ϕ ( z ) = ξ ′ ( z ) p z p − 1 ,$
then $ϕ ( z ) = 1 + b 1 z + b 2 z 2 + …$ is analytic in $U$. Using (9) and (10) we obtain
$ℜ ξ ′ ( z ) + z ξ ″ ( z ) p 2 z p − 1 = ϕ ( z ) + 1 p z ϕ ′ ( z ) > 1 − 2 ( 1 − α ) ( 1 − β ) .$
If we apply Lemma 3, then we have
$ℜ ξ ′ ( z ) p z p − 1 > 1 + 4 ( 1 − α ) ( 1 − β ) p ∑ k = 1 ∞ ( − 1 ) k p + k = : λ , ( z ∈ U ) .$
Since $ψ ( p ) > ψ ( 1 ) , p ≥ 1 ,$ it follows that $λ > 1 − 2 ( 1 − α ) ( 1 − β ) p ( 1 − ln 4 e ) .$ If
$( 1 − α ) ( 1 − β ) < p + 1 4 p 2 ( 1 − ln 4 e ) ,$
then
$λ > p − 1 2 p > 0 .$
Applying Lemma 3 again, (11) gives
$ℜ ξ ( z ) z p > 1 + 2 p ( 1 − λ ) ψ .$
If we apply Lemma 6, then we have
$ψ > − 1 p + 1 .$
Inequality (15) together with (13) implies $1 + 2 p ( 1 − λ ) ψ > 0 .$ Let $q ( z ) = z ξ ′ ( z ) p ξ ( z )$ and $τ ( z ) = ξ ( z ) z p$, then $q ( z )$ is analytic in $U$ with $q ( 0 ) = 1$ and
$ℜ { τ ( z ) } > 1 − 8 ( 1 − α ) ( 1 − β ) p 2 ψ 2 .$
By simple calculation, we find that
$ξ ′ ( z ) + z ξ ″ ( z ) p 2 z p − 1 = τ ( z ) q 2 ( z ) + 1 p z q ′ ( z ) = φ q ( z ) , z q ′ ( z ) , z ,$
where $φ ( u , v ; z ) = τ ( z ) ( u 2 + 1 p v )$. By (9) we get
$ℜ φ q ( z ) , z q ′ ( z ) , z > 1 − 2 ( 1 − α ) ( 1 − β ) ( z ∈ U ) .$
Moreover $ℜ φ ( i x , y , z ) = ℜ { τ ( z ) ( 1 p y − x 2 ) } ,$ and for real $x , y ≤ − 1 2 ( 1 + x 2 )$, we have
$ℜ φ ( i x , y , z ) ≤ − 1 2 p { 1 + ( 1 + 2 p ) x 2 } ℜ { τ ( z ) } ≤ − 1 2 p ℜ { τ ( z ) } ( z ∈ U ) .$
Thus by (16) and (17) we get
$ℜ φ ( i x , y , z ) ≤ 1 − 2 ( 1 − α ) ( 1 − β ) ,$
for all $z ∈ U$. Thus by Lemma 5, $ℜ { q ( z ) } > 0 .$ Thus, $ℜ z ξ ′ ( z ) p ξ ( z ) > 0$, that is, $ξ ∈ S p * .$
Remark 1.
Putting $p = 1$ in Theorem 1 we get the result obtained by Lashin ([26], Theorem 1).
Theorem 2.
Let $p ∈ N$ and $0 ≤ α , β , γ < 1$. If $f , g , h ∈ A p$ satisfy $f ∈ P ( p , α ) , g ∈ P ( p , β )$ and $h ∈ P ( p , γ ) ,$ then $ζ = ( f ∗ g ∗ h ) ∈ K p$, where
$( 1 − α ) ( 1 − β ) ( 1 − γ ) < min { 2 p + 1 16 p 2 ∑ k = 1 ∞ ( − 1 ) k p + k 2 + 8 p , p + 1 8 p 2 ( 1 − ln 4 e ) } .$
Proof.
It is sufficient to show that $η ( z ) = z ζ ′ ( z ) p ∈ S p * .$ Note that,
$f ′ ( z ) p z p − 1 ∗ g ′ ( z ) p z p − 1 ∗ h ′ ( z ) p z p − 1 = η ′ ( z ) + z η ″ ( z ) p 2 z p − 1 .$
By the hypothesis of Theorem 2, it follows from (18) and Lemma 4 that
$ℜ η ′ ( z ) + z η ″ ( z ) p 2 z p − 1 > 1 − 4 ( 1 − α ) ( 1 − β ) ( 1 − γ ) ,$
and the proof is completed similar to the proof of Theorem 1. □
Remark 2.
Putting $p = 1$ in Theorem 2 we get the result obtained by Lashin ([26], Theorem 2).
Theorem 3.
Let $p ∈ N , c > − p$ and $0 ≤ α < 1 .$ If $f ∈ A p$ given by (1) be in the class $P ( p , α ) ,$ then the function $J p , c$ defined by (4) belongs to the class $P ( p , β ) ,$ where
$β = 1 + 2 ( 1 − α ) ( p + c ) ∑ k = 1 ∞ ( − 1 ) k p + c + k .$
Proof.
From (4) we have
$z J p , c ″ ( z ) + ( c + 1 ) J p , c ′ ( z ) = ( c + p ) f ′ ( z ) .$
Let
$q ( z ) = J p , c ′ ( z ) p z p − 1$
so that $q ( z ) = 1 + c 1 z + c 2 z 2 + …$ is analytic in $U$. Therefore (19) and (20) leads us to
$ℜ q ( z ) + 1 p + c z q ′ ( z ) = ℜ f ′ ( z ) p z p − 1 > α ( c > − p , p ∈ N ) .$
Now by applying Lemma 3 with $λ = 1 c + p , c > − p$ and $β = α ,$ we deduce that
$ℜ J p , c ′ ( f ( z ) ) p z p − 1 > 1 + 2 ( 1 − α ) ( p + c ) ∑ k = 1 ∞ ( − 1 ) k p + c + k .$
This evidently ends the proof of Theorem 3. □
Remark 3.
The result (asserted by Theorem 3 above) was also obtained, by means of a markedly different technique, by Aouf and Ling ([33], Theorem 1).
Remark 4.
The result presented in Theorem 4 below generalizes the results shown by Ali and Thomas [14], by employing a different technique
Theorem 4.
Let $f ∈ A p$ and $J p , c$ given by (4). If $f ∈ P ( p , α ) ,$ then $J p , c ∈ S p *$$( − p < c ≤ 0 )$, where
$1 − α < min { 2 p + 1 2 ( p + c ) 1 + 2 δ ( c + p ψ ) , p + 1 2 p ( p + c ) ln 4 } ,$
$δ ( c + p ) = ∑ k = 1 ∞ ( − 1 ) k p + c + k$ and $ψ ( p ) = ∑ k = 1 ∞ ( − 1 ) k p + k .$
Proof.
Let $f ∈ A p$ be in the class $P ( p , α )$, by using Theorem 3, we have
$ℜ J p , c ′ ( f ( z ) ) p z p − 1 > 1 + 2 ( 1 − α ) ( p + c ) ∑ k = 1 ∞ ( − 1 ) k p + c + k : = μ , s a y .$
Since $δ ( c + p ) > δ ( 0 )$ for $− p < c ≤ 0 ,$ then $μ > 1 − ( 1 − α ) ( p + c ) ln 4 .$ If
$( 1 − α ) < p + 1 2 p ( p + c ) ln 4 ,$
then $μ > p − 1 2 p > 0$. Let us define the function $φ$ by
$φ ( z ) = J p , c ( z ) z p ,$
so that $φ ( z ) = 1 + c 1 z + c 2 z 2 + . . .$ is analytic in $U$ and
$ℜ φ ( z ) + 1 p z φ ′ ( z ) = ℜ J p , c ′ ( f ( z ) ) p z p − 1 > μ .$
If we apply Lemma 3 with $λ = 1 p$ and $β = μ$, then we have
$ℜ J p , c ( z ) z p > 1 + 2 p ( 1 − μ ) ψ .$
Since $2 p ( 1 − μ ) < p + 1 ,$ (15) gives $1 + 2 p ( 1 − μ ) ψ > 0 .$ Note also that from (19), we have
$z J p , c ″ ( z ) + J p , c ′ ( z ) = ( c + p ) f ′ ( z ) − c J p , c ′ ( z ) .$
Since $c ≤ 0$, the above equation and Theorem 3 give
$ℜ z J p , c ″ ( z ) + J p , c ′ ( z ) p 2 z p − 1 = ( c + p ) p ℜ f ′ ( z ) p z p − 1 − c p ℜ J p , c ′ ( z ) p z p − 1 > ( c + p ) p α − c p μ .$
Let $q ( z ) = z J p , c ′ ( z ) p J p , c ( z )$ and $ρ ( z ) = J p , c ( z ) z p ,$ then $q ( z )$ is analytic in $U$ with $q ( 0 ) = 1$ and
$ℜ { ρ ( z ) } > 1 + 2 p ( 1 − μ ) ψ .$
Applying the same method and technique as in our proof of Theorem 1, we get
$z J p , c ″ ( z ) + J p , c ′ ( z ) p 2 z p − 1 = ρ ( z ) q 2 ( z ) + 1 p z q ′ ( z ) = φ q ( z ) , z q ′ ( z ) , z ,$
where $φ ( u , v ; z ) = ρ ( z ) ( u 2 + 1 p v )$. By (23) we get
$ℜ φ q ( z ) , z q ′ ( z ) , z > ( c + p ) p α − c p μ ( z ∈ U ) .$
Moreover $ℜ φ ( i x , y , z ) = ℜ { ρ ( z ) ( 1 p y − x 2 ) } ,$ and for real $x , y ≤ − 1 2 ( 1 + x 2 )$, we have
$ℜ φ ( i x , y , z ) ≤ − 1 2 p { 1 + ( 1 + 2 p ) x 2 } ℜ { ρ ( z ) } ≤ − 1 2 p ℜ { ρ ( z ) } ( z ∈ U ) .$
Thus by (24) and (25) we get
$ℜ φ ( i x , y , z ) ≤ − 1 2 p 1 + 2 p ( 1 − μ ) ψ < ( c + p ) p α − c p μ .$
for all $z ∈ U$. Thus by Lemma 5, $ℜ { q ( z ) } > 0 .$ Thus, $ℜ z J p , c ′ ( z ) p J p , c ( z ) > 0$, that is, $J p , c ∈ S p *$ and this ends the proof. □
Remark 5.
For $p = 1$ Theorem 4 gives the result obtained by Ali and Thomas [14].
Theorem 5.
If $f ∈$$R p ( α ) ,$ then $f ∈ P ( p , α ) .$
Proof.
Let $f ∈ A p$ defined by (1) satisfies the condition (3), then
$ℜ f ′ ( z ) + z f ″ ( z ) p 2 z p − 1 = ℜ 1 + ∑ k = 1 ∞ p + k p 2 a p + k z k > α .$
Hence, we have
$ℜ 1 + 1 2 ( 1 − α ) ∑ k = 1 ∞ p + k p 2 a p + k z k > 1 2 .$
Note that
$f ′ ( z ) p z p − 1 = 1 + ∑ k = 1 ∞ p + k p a p + k z k = 1 + 1 2 ( 1 − α ) ∑ k = 1 ∞ p + k p 2 a p + k z k ∗ 1 + 2 ( 1 − α ) ∑ k = 1 ∞ p p + k z k .$
Applying Lemma 1, with $c 0 = 1$ and $c k = p p + k , k = 1 , 2 , . . . ,$ we get
$ℜ 1 + 2 ( 1 − α ) ∑ k = 1 ∞ p p + k z k > α ,$
which implies that $R e f ′ ( z ) p z p − 1 > α ,$ by using Lemma 2. □
Remark 6.
Theorem 5 is immediate from Hallenbeck-Ruscheweyh theorem [34]. Indeed, define $ϕ ( z )$ by (10) with f in place of ξ. Then $f ∈$$R p ( α )$ means $ϕ ( z ) + 1 p z ϕ ′ ( z ) ≺ 1 + ( 1 − 2 α ) z 1 − z = L ( z )$. Now Hallenbeck-Ruscheweyh theorem (see also Miller-Mocanu ([29], P.71, Theorem 3.1b)) implies $ϕ ( z ) ≺ L ( z )$,
Remark 7.
Putting $p = 1$ in Theorem 5 we get the result obtained by Al-Oboudi ([35], Theorem 2.3, when $λ = n = 1$).
Theorem 6.
Let $f ∈$$R p ( α ) .$ Then $f ∈$$P ( p , ξ ) ,$ where
$ξ = 2 + ( p 2 + 3 p ) α ( 1 + p ) ( p + 2 ) ≥ α .$
Proof.
It is shown in [36] that, if $γ ≥ 0$ and if $g ( z ) = z + ∑ k = 1 ∞ 1 1 + γ k z k + 1$ then
$ℜ g ( z ) z ≥ 2 γ 2 + 3 γ + 1 2 ( 1 + γ ) ( 1 + 2 γ ) .$
Hence
$ℜ 1 + 2 ( 1 − α ) ∑ k = 1 ∞ p p + k z k ≥ ( 2 + p 2 + 3 p ) α ( 1 + p ) ( p + 2 ) .$
Using (26) in the Theorem 5 we get the result. □
Remark 8.
Putting $p = 1$ in Theorem 6 we get the result obtained by Al-Oboudi ([35], Remark 2.5, when $λ = 1$).

## 4. Conclusions

The convolution method has recently been used to study many interesting subclasses of analytical functions. An interesting criterion was given by Lashin [26] to be starlike for convolution of functions with positive real parts, which was improved by Sokol [27]. Each of these earlier results has been extended and improved in this paper. Additionally, by using Miller and Mocanu Theorem [29], Ali and Thomas’ results [14] for the starlikeness of the Bernardi integral operator have been extended.

## Author Contributions

Conceptualization, A.M.Y.L.; Funding acquisition, A.M.Y.L.; Investigation, A.M.Y.L. and M.K.A.; Project administration, A.M.Y.L.; Supervision, A.M.Y.L. and M.K.A.; Writing—original draft, A.M.Y.L. and M.K.A.; Writing—review and editing, A.M.Y.L. and M.K.A. All authors have read and agreed to the published version of the manuscript.

## Funding

The Deanship of Scientific Research (DSR) at King Abdulaziz University (KAU), Jeddah, Saudi Arabia has funded this project, under grant no. (G: 319-130-1443).

Not applicable.

Not applicable.

Not applicable.

## Acknowledgments

The authors, acknowledge with thanks DSR for technical and financial support. Also the authors would like to express their thanks to the referees for their helpful comments and suggestions that improved the presentation of the paper.

## Conflicts of Interest

The authors declare no conflict of interest.

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Lashin, A.M.Y.; Aouf, M.K. Hadamard Product of Certain Multivalent Analytic Functions with Positive Real Parts. Mathematics 2022, 10, 1506. https://doi.org/10.3390/math10091506

AMA Style

Lashin AMY, Aouf MK. Hadamard Product of Certain Multivalent Analytic Functions with Positive Real Parts. Mathematics. 2022; 10(9):1506. https://doi.org/10.3390/math10091506

Chicago/Turabian Style

Lashin, Abdel Moneim Y., and Mohamed K. Aouf. 2022. "Hadamard Product of Certain Multivalent Analytic Functions with Positive Real Parts" Mathematics 10, no. 9: 1506. https://doi.org/10.3390/math10091506

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