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Article

# Subordination and Superordination Properties for Certain Family of Analytic Functions Associated with Mittag–Leffler Function

by
Mansour F. Yassen
1,2,
3,4 and
Praveen Agarwal
5,6,7,8,*
1
Department of Mathematics, College of Science and Humanities in Al-Aflaj, Prince Sattam Bin Abdulaziz University, Al-Aflaj 11912, Saudi Arabia
2
Department of Mathematics, Faculty of Science, Damietta University, New Damietta 34517 Damietta, Egypt
3
Department of Mathematics, College of Science, University of Ha’il, Ha’il 81451, Saudi Arabia
4
Department of Mathematics, Faculty of Science, University of Mansoura, Mansoura 35516, Egypt
5
Department of Mathematics, Anand International College of Engineering, Jaipur 303012, India
6
International Center for Basic and Applied Sciences, Jaipur 302029, India
7
Institute of Mathematics and Mathematical Modeling, Almaty 050010, Kazakhstan
8
An Aided Institute of the Department of Atomic Energy, Harish-Chandra Research Institute (HRI), Allahabad 211 019, India
*
Author to whom correspondence should be addressed.
Symmetry 2020, 12(10), 1724; https://doi.org/10.3390/sym12101724
Submission received: 24 August 2020 / Revised: 9 September 2020 / Accepted: 9 October 2020 / Published: 19 October 2020

## Abstract

:
We obtain new outcomes of analytic functions linked with operator $H α , β η , k ( f )$ defined by Mittag–Leffler function. Moreover, new theorems of differential sandwich-type are obtained.

## 1. Basic Definitions and Preliminaries

Let $A$ define the class of analytic functions in the open unit disk $U = { z ∈ C : | z | < 1 }$ and let $H [ a , n ]$ be the subclass of $A$, which is
$f ( z ) = a + a n z n + a n + 1 z n + 1 + a n + 2 z n + 2 + a n + 3 z n + 3 + … ( a ∈ C ) ,$
Furthermore, let $H$ be the subclass of $A$ of all the functions $f ( z ) ∈ H$ normalized by
$f ( z ) = z + ∑ n = 2 ∞ a n z n .$
Attiya [1] introduced and investigated the operator $H α , β η , k ( f ( z ) ) : H → H$, which $H α , β η , k ( f ( z ) )$ is defined by
$H α , β η , k ( f ( z ) ) = Q α , β η , k ( z ) ∗ f ( z ) , ( z ∈ U ) ,$
for $f ( z ) ∈ H$ given by (1), the symbol ∗ denotes the Hadamard product, and
$Q α , β η , k ( z ) = Γ ( α + β ) ( η ) k E α , β η , k ( z ) − 1 Γ ( β ) , ( z ∈ U ) .$
Moreover, the function $E α , β η , k ( z )$ is called the general Mittag–Leffler function defined by
$E α , β η , k ( z ) = ∑ n = 0 ∞ ( η ) n k z n Γ ( α n + β ) n ! , ( α , β , η ∈ C ; R e ( α ) > m a x { 0 , R e ( k ) − 1 } ; R e ( k ) > 0 ) ,$
where
$( η ) n = Γ ( η + n ) Γ ( η ) = 1 , n = 0 , η ( η + 1 ) ( η + 2 ) … ( η + n − 1 ) , n ∈ N .$
The function $E α , β η , k ( z )$ was investigated by Srivastava and Tomovski [2]. Many authors studied and investigated Mittag–Leffler function; for more details on Mittag–Leffler function and general Mittag–Leffler function see, e.g., [1,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19].
Moreover, Attiya [1] deduced that $H α , β η , k ( f ( z ) )$ can be put in
$H α , β η , k ( f ( z ) ) = z + ∑ n = 2 ∞ Γ ( η + n k ) Γ ( α + β ) Γ ( η + k ) Γ ( n α + β ) a n z n ( z ∈ U ) .$
It follows from (2) that (see [1])
$k z H α , β η , k ( f ( z ) ) ′ = ( η + k ) H α , β η + 1 , k ( f ( z ) ) − η H α , β η , k ( f ( z ) ) ,$
and
$α z H α , β + 1 η , k ( f ( z ) ) ′ = ( α + β ) H α , β η , k ( f ( z ) ) − β H α , β + 1 η , k ( f ( z ) ) .$
It should be remarked that the operator $H α , β η , k ( f ( z ) )$ for some special cases of $α , β , η$, and k provides many special functions, e.g.,
$H 0 , β 1 , 1 ( f ) ( z ) = f ( z ) .$
$H 0 , β 2 , 1 ( f ) ( z ) = 1 2 ( f ( z ) + z f ′ ( z ) ) .$
$H 0 , β 0 , 1 ( f ) ( z ) = ∫ 0 z 1 t f ( t ) d t .$
$H 1 , 0 1 , 1 ( z 1 − z ) = z e z .$
$H 1 , 1 1 , 1 ( z 1 − z ) = e z − 1 .$
$H 2 , 1 1 , 1 ( z 1 − z ) = − 2 + cosh ( z ) .$
Definition 1.
Let functions $f ( z )$ and $g ( z )$ be analytic in the open unit disk $U$. Then $f ( z )$ is subordinate to $g ( z )$ if there exists a Schwarz function $ω ( z )$, analytic in $U$ with $ω ( 0 ) = 0$ and $| ω ( z ) | < 1 , ( z ∈ U )$, such that $f ( z ) = g ( ω ( z ) ) , ( z ∈ U )$, we denote this subordination by $f ( z ) ≺ g ( z )$. In particular, if $g ( z )$ is univalent in $U$, then subordination is equivalent to $f ( z ) ≺ g ( z ) ⇔ f ( 0 ) = g ( 0 )$ and $f ( U ) ⊂ g ( U )$.
Definition 2.
If $Q$ the set of all functions $q ( z )$ that are analytic and univalent on $U ¯ \ E ( q )$, where
$E ( q ) = ξ ∈ ∂ U : lim z → ξ q ( z ) = ∞ ,$
and $m i n | q ′ ( ξ ) | = ρ > 0$ for $ξ ∈ ∂ U \ E ( q )$. Further, let $Q ( a ) = q ( z ) ∈ U : q ( 0 ) = a$ and $Q 1 = Q ( 1 )$.
Definition 3.
If $ψ : C 4 × U ⟶ C$ and $h ( z )$ be univalent in $U$. If $p ( z )$ is analytic in $U$, and satisfies the third-order differential subordination
$ψ ( p ( z ) , z p ′ ( z ) , z 2 p ″ ( z ) , z 3 p ‴ ( z ) ; z ) ≺ h ( z ) ,$
then $p ( z )$ is called a solution of the differential subordination and $q ( z )$ is called a dominant of the solutions of the differential subordination as well as a dominant if $p ( z ) ≺ q ( z )$ for all $p ( z )$ satisfying (5). $q ˜ ( z )$ that satisfies $q ˜ ( z ) ≺ q ( z )$ for all dominants of (5) is called the best dominant of (5).
Definition 4.
Let $Ω ⊆ C$, $q ( z ) ∈ Q$ and $n ∈ N \ { 1 }$. The class of admissible functions $Ψ n [ Ω , q ( z ) ]$ consists of those functions $ψ : C 4 × U ⟶ C$ that satisfy the admissibility condition:
$ψ ( r , s , t , u ; z ) ∉ Ω ,$
whenever
$r = q ( ζ ) , s = ℓ ζ q ′ ( ζ ) , R e t s + 1 ≥ ℓ R e ζ q ″ ( ζ ) q ′ ( ζ ) + 1 ,$
and
$R e u s ≥ ℓ 2 R e ζ 2 q ‴ ( ζ ) q ′ ( ζ ) ,$
where $z ∈ U$; $ζ ∈ ∂ U \ E ( q )$ and $ℓ ≥ n$.
Analogous to the second order differential super-ordinations introduced by Miller and Mocanu [20], Tang et al. [21] defined the differential super-ordinations as follows:
Definition 5.
Let $ψ : C 4 × U ⟶ C$ and the function $h ( z )$ be analytic in $U$. If functions $p ( z )$ and $ψ ( p ( z ) , z p ′ ( z ) , z 2 p ″ ( z ) , z 3 p ‴ ( z ) )$ are univalent in $U$, and satisfy the following third-order differential super-ordination
$h ( z ) ≺ ψ ( p ( z ) , z p ′ ( z ) , z 2 p ″ ( z ) , z 3 p ‴ ( z ) ) ,$
then $p ( z )$ is called a solution of the differential superordination and $q ( z )$ is called a subordinant of the solutions of the differential super-ordinations as well as a subordinant if $p ( z ) ≺ q ( z )$ for all $p ( z )$ satisfying Equation (6). A univalent subordinant $q ˜ ( z )$ that satisfies $q ˜ ( z ) ≺ q ( z )$ for all super-ordinations of (6) is the best superordinant.
Definition 6.
Let $Ω ⊆ C$, $q ( z ) ∈ H [ a , n ]$ with $n ∈ N \ { 1 }$ and $q ′ ( z ) ≠ 0$. The class of admissible functions $Ψ n ′ [ Ω , q ( z ) ]$ consists of those functions $ψ : C 4 × U ⟶ C$ that satisfy the admissibility condition:
$ψ ( r , s , t , u ; ζ ) ∈ Ω ,$
whenever
$r = q ( z ) , s = z q ′ ( z ) m , R e t s + 1 ≤ 1 m R e z q ″ ( z ) q ′ ( z ) + 1 ,$
and
$R e u s ≤ 1 m 2 R e z 2 q ‴ ( z ) q ′ ( z ) ,$
where $z ∈ U$; $ζ ∈ ∂ U$ and $m ≥ n ≥ 2$.
Here, we use the following theorems given by Antonino and Miller [22]:
Theorem 1
([22]). Let $p ( z ) ∈ H [ a , n ]$ with $n ∈ N \ { 1 }$. Also, let $q ( z ) ∈ Q ( a )$ and satisfy the following conditions:
$R e ζ q ″ ( ζ ) q ′ ( ζ ) > 0 , z p ′ ( z ) q ′ ( ζ ) ≤ ℓ , ( z ∈ U ; ζ ∈ ∂ U \ E ( q ) ; ℓ ≥ n ) , ψ ( p ( z ) , z p ′ ( z ) , z 2 p ″ ( z ) , z 3 p ‴ ( z ) ; z ) ∈ Ω ,$
and, if $Ω ⊆ C$, $ψ ∈ Ψ n [ Ω , q ( z ) ]$, then $p ( z ) ≺ q ( z )$.
Theorem 2.
Let $q ( z ) ∈ H [ a , n ]$ and $ψ ∈ Ψ n ′ [ Ω , q ( z ) ]$. If $p ( z ) ∈ Q ( a )$ and $ψ ( p ( z ) , z p ′ ( z ) , z 2 p ″ ( z ) , z 3 p ‴ ( z ) ; z )$ is univalent in $U$ and
$R e z q ″ ( z ) q ′ ( z ) ≥ 0 , ζ p ′ ( ζ ) q ′ ( z ) ≤ m , ( z ∈ U ; ζ ∈ ∂ U ; m ≥ n ≥ 2 ) Ω ⊂ ψ ( p ( z ) , z p ′ ( z ) , z 2 p ″ ( z ) , z 3 p ‴ ( z ) ; z ) : z ∈ U ,$
then $q ( z ) ≺ p ( z )$.
Here, we study a certain family of admissible functions by using the third-order differential subordination and superordination given by Antonino and Miller [22] and Tang et al. [21]—see also Attiya et al. [23]—we obtain new results of subordination and superordination properties of analytic functions linked with the operator $H α , β η , k ( f )$.

## 2. Main Results

Definition 7.
Let $Ω ⊆ C$ and $q ( z ) ∈ Q$. The class of admissible functions $Ψ Γ [ Ω , q ( z ) ]$ consists of those functions $ϕ : C 4 × U ⟶ C$ that satisfy the admissibility condition
$ϕ ( a 1 , a 2 , a 3 , a 4 ; z ) ∉ Ω ,$
whenever
$a 1 = q ( ζ ) , a 2 = ℓ k ζ q ′ ( ζ ) + b q ( ζ ) b ,$
$R e ( b + 1 ) ( a 3 − a 1 ) k ( a 2 − a 1 ) − 2 b + 1 k ≥ ℓ R e ζ q ″ ( ζ ) q ′ ( ζ ) + 1 ,$
$R e ( ( b + 1 ) ( b + 2 ) ( a 4 − a 1 ) − 3 ( b + 1 ) ( b + k + 1 ) ( a 3 − a 1 ) k 2 ( a 2 − a 1 ) + 3 b ( b + 1 ) + 1 k 2 + 6 b + 3 k + 2 ) ≥ ℓ 2 R e ζ 2 q ‴ ( ζ ) q ′ ( ζ ) ,$
where $z ∈ U$; $ζ ∈ ∂ U \ E ( q )$, $ℓ ∈ N \ { 1 }$ and $b = η + k$.
Theorem 3.
Let $ϕ ∈ Ψ Γ [ Ω , q ( z ) ]$. If $f ( z ) ∈ H$ and $q ( z ) ∈ Q 1$ satisfy:
$R e ζ q ″ ( ζ ) q ′ ( ζ ) ≥ 0 , H α , β η + 1 , k ( f ( z ) ) − H α , β η , k ( f ( z ) ) z q ′ ( ζ ) ≤ | k b | ℓ ,$
$ϕ H α , β η , k ( f ( z ) ) z , H α , β η + 1 , k ( f ( z ) ) z , H α , β η + 2 , k ( f ( z ) ) z , H α , β η + 3 , k ( f ( z ) ) z ; z : z ∈ U ⊂ Ω ,$
then
$H α , β η , k ( f ( z ) ) ≺ q ( z ) .$
Proof.
Let
$H α , β η , k ( f ( z ) ) = z p ( z ) z ∈ U ,$
From (3), we have
$H α , β η + 1 , k ( f ( z ) ) z = ( k b ) z p ′ ( z ) + b k p ( z ) ,$
which implies
$H α , β η + 2 , k ( f ( z ) ) z = k 2 b ( b + 1 ) z 2 p ″ ( z ) + ( 2 b + 1 k + 1 ) z p ′ ( z ) + b ( b + 1 ) k 2 p ( z ) .$
Furthermore, we have
$H α , β η + 3 , k ( f ( z ) ) z = k 3 b ( b + 1 ) ( b + 2 ) ( z 3 p ‴ ( z ) + 3 ( b + 1 k + 1 ) z 2 p ″ ( z ) + ( 3 b 2 + 6 b + 2 k 2 + 3 ( b + 1 ) k + 1 ) z p ′ ( z ) + b ( b + 1 ) ( b + 2 ) k 3 p ( z ) ) .$
Now, we define the parameters $a 1 , a 2 , a 3$, and $a 4$ as
$a 1 = r , a 2 = ( k b ) s + b k r , a 3 = k 2 b ( b + 1 ) t + ( 2 b + 1 k + 1 ) s + b ( b + 1 ) k 2 r ,$
and
$a 4 = k 3 b ( b + 1 ) ( b + 2 ) ( u + 3 ( b + 1 k + 1 ) t + ( 3 b 2 + 6 b + 2 k 2 + 3 ( b + 1 ) k + 1 ) s + b ( b + 1 ) ( b + 2 ) k 3 r ) .$
Then, transformation $ψ : C 4 × U ⟶ C$ as
$ψ ( r , s , t , u ; z ) = ϕ ( a 1 , a 2 , a 3 , a 4 ; z ) ,$
by using the relations from (9) to (12), we have
$ψ ( p ( z ) , z p ′ ( z ) , z 2 p ″ ( z ) , z 3 p ‴ ( z ) ; z ) = ϕ ( H α , β η , k ( f ( z ) ) z , H α , β η + 1 , k ( f ( z ) ) z , H α , β η + 2 , k ( f ( z ) ) z , H α , β η + 3 , k ( f ( z ) ) z ; z ) ,$
therefore, we recompute (8) as
$ψ ( p ( z ) , z p ′ ( z ) , z 2 p ″ ( z ) , z 3 p ‴ ( z ) ; z ) ∈ Ω ,$
then, the proof is completed by showing that the admissibility condition for $ϕ ∈ Ψ Γ [ Ω , q ( z ) ]$ is equivalent to the admissibility condition for $ψ$ as given in Definition 3, since
$t s + 1 = ( b + 1 ) ( a 3 − a 1 ) k ( a 2 − a 1 ) − 2 b + 1 k ,$
and
$u s = ( b + 1 ) ( b + 2 ) ( a 4 − a 1 ) − 3 ( b + 1 ) ( b + k + 1 ) ( a 3 − a 1 ) k 2 ( a 2 − a 1 ) + 3 b ( b + 1 ) + 1 k 2 + 6 b + 3 k + 2 ,$
we also note that
$z p ′ ( z ) q ′ ( ζ ) = | ( b z k ) H α , β η + 1 , k ( f ( z ) ) − H α , β η , k ( f ( z ) ) q ′ ( ζ ) | ≤ ℓ ,$
therefore, $ψ ∈ Ψ Γ [ Ω , q ( z ) ]$ and by Theorem 1, $p ( z ) ≺ q ( z )$. □
In a similar way, we define the parameters $a 1 , a 2 , a 3$, and $a 4$ as follows:
Definition 8.
Let $Ω ⊆ C$ and $q ( z ) ∈ Q$. The class of admissible functions $Ψ Γ [ Ω , q ( z ) ]$ consists of those functions $ϕ : C 4 × U ⟶ C$ that satisfy the admissibility condition
$ϕ ( a 1 , a 2 , a 3 , a 4 ; z ) ∉ Ω ,$
whenever
$a 1 = q ( ζ ) , a 2 = ℓ α ζ q ′ ( ζ ) + c q ( ζ ) c ,$
$R e ( c − 1 ) ( a 3 − a 1 ) α ( a 2 − a 1 ) − 2 c − 1 α ≥ ℓ R e ζ q ″ ( ζ ) q ′ ( ζ ) + 1 ,$
$R e ( ( c − 1 ) ( c − 2 ) ( a 4 − a 1 ) − 3 ( c − 1 ) ( c + α − 1 ) ( a 3 − a 1 ) α 2 ( a 2 − a 1 ) + 3 c ( c − 1 ) + 1 α 2 + 6 c − 3 α + 2 ) ≥ ℓ 2 R e ζ 2 q ‴ ( ζ ) q ′ ( ζ ) ,$
where $z ∈ U$; $ζ ∈ ∂ U \ E ( q )$, $ℓ ∈ N \ { 1 }$ and $c = α + β$.
Theorem 4.
Let $ϕ ∈ Ψ Γ [ Ω , q ( z ) ]$. If $f ( z ) ∈ H$ and $q ( z ) ∈ Q 1$ satisfy the following conditions:
$R e ζ q ″ ( ζ ) q ′ ( ζ ) ≥ 0 , 1 z H α , β η , k ( f ( z ) ) − H α , β + 1 η , k ( f ( z ) ) q ′ ( ζ ) ≤ | α c | ℓ ,$
$ϕ H α , β + 1 η , k ( f ( z ) ) z , H α , β η , k ( f ( z ) ) z , H α , β − 1 η , k ( f ( z ) ) z , H α , β − 2 η , k ( f ( z ) ) z ; z : z ∈ U ⊂ Ω ,$
then
$H α , β + 1 η , k ( f ( z ) ) ≺ q ( z ) .$
Proof.
Let
$H α , β + 1 η , k ( f ( z ) ) = z p ( z ) z ∈ U ,$
From (4), we have
$H α , β η , k ( f ( z ) ) z = ( α c ) z p ′ ( z ) + c α p ( z ) ,$
which implies
$H α , β − 1 η , k ( f ( z ) ) z = α 2 c ( c − 1 ) z 2 p ″ ( z ) + ( 2 c − 1 α + 1 ) z p ′ ( z ) + c ( c − 1 ) α 2 p ( z ) .$
Moreover, we have
$H α , β − 2 η , k ( f ( z ) ) z = α 3 c ( c − 1 ) ( c − 2 ) ( z 3 p ‴ ( z ) + 3 ( c − 1 α + 1 ) z 2 p ″ ( z ) + ( 3 c 2 − 6 c + 2 α 2 + 3 ( c − 1 ) α + 1 ) z p ′ ( z ) + c ( c − 1 ) ( c − 2 ) α 3 p ( z ) ) .$
Parameters $a 1 , a 2 , a 3$ and, $a 4$ as
$a 1 = r , a 2 = ( α c ) s + c α r , a 3 = α 2 c ( c − 1 ) t + ( 2 c − 1 α + 1 ) s + c ( c − 1 ) α 2 r ,$
and
$a 4 = α 3 c ( c − 1 ) ( c − 2 ) ( u + 3 ( c − 1 α + 1 ) t + ( 3 c 2 − 6 c + 2 α 2 + 3 ( c − 1 ) α + 1 ) s + c ( c − 1 ) ( c − 2 ) α 3 r ) .$
The transformation $ψ : C 4 × U ⟶ C$
$ψ ( r , s , t , u ; z ) = ϕ ( a 1 , a 2 , a 3 , a 4 ; z ) ,$
by using the relations from (17) to (20), we have
$ψ ( p ( z ) , z p ′ ( z ) , z 2 p ″ ( z ) , z 3 p ‴ ( z ) ; z ) = ϕ ( H α , β + 1 η , k ( f ( z ) ) z , H α , β η , k ( f ( z ) ) z , H α , β − 1 η + 2 , k ( f ( z ) ) z , H α , β − 2 η , k ( f ( z ) ) z ; z ) ,$
we recompute (16) as
$ψ ( p ( z ) , z p ′ ( z ) , z 2 p ″ ( z ) , z 3 p ‴ ( z ) ; z ) ∈ Ω ,$
This completes the proof by showing that the admissibility condition for $ϕ ∈ Ψ Γ [ Ω , q ( z ) ]$ is equivalent to the admissibility condition for $ψ$ as given in Definition 3, since
$t s + 1 = ( c − 1 ) ( a 3 − a 1 ) α ( a 2 − a 1 ) − 2 c − 1 α ,$
and
$u s = ( c − 1 ) ( c − 2 ) ( a 4 − a 1 ) − 3 ( c − 1 ) ( c + α − 1 ) ( a 3 − a 1 ) α 2 ( a 2 − a 1 ) + 3 c ( c − 1 ) + 1 α 2 + 6 c − 3 α + 2 ,$
we also note that
$z p ′ ( z ) q ′ ( ζ ) = ( c z α ) H α , β η , k ( f ( z ) ) − H α , β + 1 η , k ( f ( z ) ) q ′ ( ζ ) ≤ ℓ ,$
therefore, $ψ ∈ Ψ Γ [ Ω , q ( z ) ]$ and hence by Theorem 1, $p ( z ) ≺ q ( z )$. □
If $Ω ≠ C$ is simply connected to the domain, then $Ω = h ( U )$ for some conformal mapping $h ( z )$ of $U$ onto $Ω$. In this case, the class $Ψ Γ [ h ( U ) , q ( z ) ]$ is written as $Ψ Γ [ h , q ]$; the following theorem is a direct consequence of Theorems 3 and 4.
Theorem 5.
Let $ϕ ∈ Ψ Γ [ h , q ]$. If $f ( z ) ∈ H$ and $q ( z ) ∈ Q 1$ satisfy the following conditions:
$( i ) R e ζ q ″ ( ζ ) q ′ ( ζ ) ≥ 0 , 1 z H α , β η + 1 , k ( f ( z ) ) − H α , β η , k ( f ( z ) ) q ′ ( ζ ) ≤ | k b | ℓ ,$
$( i i ) ϕ H α , β η , k ( f ( z ) ) z , H α , β η + 1 , k ( f ( z ) ) z , H α , β η + 2 , k ( f ( z ) ) z , H α , β η + 3 , k ( f ( z ) ) z ; z ≺ h ( z ) ,$
then
$H α , β η , k ( f ( z ) ) ≺ q ( z ) .$
Theorem 6.
Let $ϕ ∈ Ψ Γ [ h , q ]$. If $f ( z ) ∈ H$ and $q ( z ) ∈ Q 1$ satisfy the following conditions:
$R e ζ q ″ ( ζ ) q ′ ( ζ ) ≥ 0 , 1 z H α , β η , k ( f ( z ) ) − H α , β + 1 η , k ( f ( z ) ) q ′ ( ζ ) ≤ | α c | ℓ ,$
$ϕ H α , β + 1 η , k ( f ( z ) ) z , H α , β η , k ( f ( z ) ) z , H α , β − 1 η , k ( f ( z ) ) z , H α , β − 2 η , k ( f ( z ) ) z ; z ≺ h ( z ) ,$
then
$H α , β + 1 η , k ( f ( z ) ) ≺ q ( z ) .$
The next corollaries extend Theorems 3 and 4, when the behavior of $q ( z )$ on $∂ U$ is not known.
Corollary 1.
Let $Ω ⊂ C$ and let $q ( z )$ be univalent in $U$; $q ( 0 ) = 1$. Let $ϕ ∈ Ψ Γ [ Ω , q σ ]$ for some $σ ∈ ( 0 , 1 )$ where $q σ ( z ) = q ( σ z )$. If $f ( z ) ∈ H$ satisfies the following conditions:
$R e ζ q σ ″ ( ζ ) q σ ′ ( ζ ) ≥ 0 , 1 z H α , β η + 1 , k ( f ( z ) ) − H α , β η , k ( f ( z ) ) q σ ′ ( ζ ) ≤ | k b | ℓ ,$
$ϕ H α , β η , k ( f ( z ) ) z , H α , β η + 1 , k ( f ( z ) ) z , H α , β η + 2 , k ( f ( z ) ) z , H α , β η + 3 , k ( f ( z ) ) z ; z ∈ Ω ,$
then
$H α , β η , k ( f ( z ) ) ≺ q ( z ) ,$
where $z ∈ U$ and $ζ ∈ ∂ U \ E ( q σ ) .$
Proof.
by using Theorem 3, we have $H α , β η , k ( f ( z ) ) ≺ q σ ( z )$. Then we obtain the result from $q σ ( z ) ≺ q ( z )$. □
Corollary 2.
Let $Ω ⊂ C$ and let $q ( z )$ be univalent in $U$; $q ( 0 ) = 1$. Let $ϕ ∈ Ψ Γ [ Ω , q σ ]$ for some $σ ∈ ( 0 , 1 )$, where $q σ ( z ) = q ( σ z )$. If $f ( z ) ∈ H$ satisfy the following conditions:
$R e ζ q σ ″ ( ζ ) q σ ′ ( ζ ) ≥ 0 , 1 z H α , β η , k ( f ( z ) ) − H α , β + 1 η , k ( f ( z ) ) q σ ′ ( ζ ) ≤ | α c | ℓ ,$
$ϕ H α , β + 1 η , k ( f ( z ) ) z , H α , β η , k ( f ( z ) ) z , H α , β − 1 η , k ( f ( z ) ) z , H α , β − 2 η , k ( f ( z ) ) z ; z ∈ Ω ,$
then
$H α , β + 1 η , k ( f ( z ) ) ≺ q ( z ) ,$
where $z ∈ U$ and $ζ ∈ ∂ U \ E ( q σ ) .$
Proof.
By using Theorem 4, we have $H α , β + 1 η , k ( f ( z ) ) ≺ q σ ( z )$. Then we obtain the result from $q σ ( z ) ≺ q ( z )$. □
Corollary 3.
Let $Ω ⊂ C$ and let $q ( z )$ be univalent in $U$; $q ( 0 ) = 1$. Let $ϕ ∈ Ψ Γ [ Ω , q σ ]$ for some $σ ∈ ( 0 , 1 )$, where $q σ ( z ) = q ( σ z )$. If $f ( z ) ∈ H$ satisfy the following conditions:
$R e ζ q σ ″ ( ζ ) q σ ′ ( ζ ) ≥ 0 , 1 z H α , β η + 1 , k ( f ( z ) ) − H α , β η , k ( f ( z ) ) q σ ′ ( ζ ) ≤ | k b | ℓ ,$
$ϕ H α , β η , k ( f ( z ) ) z , H α , β η + 1 , k ( f ( z ) ) z , H α , β η + 2 , k ( f ( z ) ) z , H α , β η + 3 , k ( f ( z ) ) z ; z ≺ h ( z ) ,$
then
$H α , β η , k ( f ( z ) ) ≺ q ( z ) ,$
where $z ∈ U$ and $ζ ∈ ∂ U \ E ( q σ ) .$
Corollary 4.
Let $Ω ⊂ C$ and let $q ( z )$ be univalent in $U$; $q ( 0 ) = 1$. Let $ϕ ∈ Ψ Γ [ Ω , q σ ]$ for some $σ ∈ ( 0 , 1 )$, where $q σ ( z ) = q ( σ z )$. If $f ( z ) ∈ H$ satisfy the following conditions:
$R e ζ q σ ″ ( ζ ) q σ ′ ( ζ ) ≥ 0 , 1 z H α , β η , k ( f ( z ) ) − H α , β + 1 η , k ( f ( z ) ) q σ ′ ( ζ ) ≤ | α c | ℓ ,$
$ϕ H α , β + 1 η , k ( f ( z ) ) z , H α , β η , k ( f ( z ) ) z , H α , β − 1 η , k ( f ( z ) ) z , H α , β − 2 η , k ( f ( z ) ) z ; z ≺ h ( z ) ,$
then
$H α , β + 1 η , k ( f ( z ) ) ≺ q ( z ) ,$
where $z ∈ U$ and $ζ ∈ ∂ U \ E ( q σ ) .$
Theorem 7.
Let $h ( z )$ be univalent in $U$. Let $ϕ : C 4 × U ⟶ C$. Suppose that the differential equation
$ϕ ( q ( z ) , ( k b ) z q ′ ( z ) + b k q ( z ) , k 2 b ( b + 1 ) z 2 q ″ ( z ) + ( 2 b + 1 k + 1 ) z q ′ ( z ) + b ( b + 1 ) k 2 q ( z ) , k 3 b ( b + 1 ) ( b + 2 ) ( z 3 q ‴ ( z ) + 3 ( b + 1 k + 1 ) z 2 q ″ ( z ) + ( 3 b 2 + 6 b + 2 k 2 + 3 ( b + 1 ) k + 1 ) z q ′ ( z ) + b ( b + 1 ) ( b + 2 ) k 3 q ( z ) ) ; z ) = h ( z ) ,$
has a solution $q ( z )$ with $q ( 0 ) = 1$, which satisfies (7). If $f ( z ) ∈ H$ satisfies (24) and
$ϕ H α , β η , k ( f ( z ) ) z , H α , β η + 1 , k ( f ( z ) ) z , H α , β η + 2 , k ( f ( z ) ) z , H α , β η + 3 , k ( f ( z ) ) z ; z ,$
is analytic in $U$, then
$H α , β η , k ( f ( z ) ) ≺ q ( z ) ,$
and $q ( z )$ is the best dominant of (27).
Proof.
By using Theorem 3 that $q ( z )$ is a dominant of (24). Since $q ( z )$ satisfies (26), it is also a solution of (24) and therefore $q ( z )$ will be dominated by all dominants. Hence, $q ( z )$ is the best dominant. □
Moreover, in a similar way, using Theorem 4, we have
Theorem 8.
Let $h ( z )$ be univalent in $U$. Let $ϕ : C 4 × U ⟶ C$. Suppose that the differential equation
$ϕ ( q ( z ) , ( α c ) z q ′ ( z ) + c α q ( z ) , α 2 c ( c − 1 ) z 2 q ″ ( z ) + ( 2 c − 1 α + 1 ) z q ′ ( z ) + c ( c − 1 ) α 2 q ( z ) , α 3 c ( c − 1 ) ( c − 2 ) ( z 3 q ‴ ( z ) + 3 ( c − 1 α + 1 ) z 2 q ″ ( z ) + ( 3 c 2 − 6 c + 2 α 2 + 3 ( c − 1 ) α + 1 ) z q ′ ( z ) + c ( c − 1 ) ( c − 2 ) α 3 q ( z ) ) ; z ) = h ( z ) ,$
has a solution $q ( z )$ with $q ( 0 ) = 1$, which satisfies (15). If $f ( z ) ∈ H$ satisfies (25) and
$ϕ H α , β + 1 η , k ( f ( z ) ) z , H α , β η , k ( f ( z ) ) z , H α , β − 1 η , k ( f ( z ) ) z , H α , β − 2 η , k ( f ( z ) ) z ; z ,$
is analytic in $U$, then
$H α , β + 1 η , k ( f ( z ) ) ≺ q ( z ) .$
and $q ( z )$ is the best dominant of (29).
In the case $q ( z ) = 1 + M z , ( M > 0 )$ and in view of the Definition 7, the class of admissible functions $Ψ Γ [ Ω , q ]$ denoted by $Ψ Γ [ Ω , M ]$ is defined below.
Definition 9.
Let $Ω ⊆ C$ and $M > 0$. The class of admissible functions $Ψ Γ [ Ω , M ]$ consists of those functions $ϕ : C 4 × U ⟶ C$ that satisfy the admissibility condition
$ϕ ( 1 + M e i θ , 1 + ( k b ) b k + ℓ M e i θ , 1 + k 2 b ( b + 1 ) L + b ( b + 1 ) k 2 + ℓ ( 2 b + 1 k + 1 ) M e i θ , 1 + k 3 b ( b + 1 ) ( b + 2 ) ( N + 3 L ( b + 1 k + 1 ) + ( b ( b + 1 ) ( b + 2 ) k 3 + ℓ ( 3 b 2 + 6 b + 2 k 2 + 3 ( b + 1 ) k + 1 ) ) M e i θ ) ; z ) ∉ Ω ,$
where $z ∈ U$, $R e ( L e − i θ ) ≥ ℓ ( ℓ − 1 ) M$ and $R e ( N e − i θ ) ≥ 0$ for all real θ and $ℓ ∈ N \ { 1 }$.
Corollary 5.
Let $ϕ ∈ Ψ Γ [ Ω , M ]$. If $f ( z ) ∈ H$ satisfy the following conditions:
$1 z H α , β η + 1 , k ( f ( z ) ) − H α , β η , k ( f ( z ) ) ≤ | k b | ℓ M ,$
and
$ϕ H α , β η , k ( f ( z ) ) z , H α , β η + 1 , k ( f ( z ) ) z , H α , β η + 2 , k ( f ( z ) ) z , H α , β η + 3 , k ( f ( z ) ) z ; z ∈ Ω ,$
then
$H α , β η , k ( f ( z ) ) z − 1 < M .$
Furthermore, with Definition 8, we can define the following:
Definition 10.
Let $Ω ⊆ C$ and $M > 0$. The class of admissible functions $Ψ Γ [ Ω , M ]$ consists of those functions $ϕ : C 4 × U ⟶ C$ that satisfy the admissibility condition
$ϕ ( 1 + M e i θ , 1 + ( α c ) c α + ℓ M e i θ , 1 + α 2 c ( c − 1 ) L + c ( c − 1 ) α 2 + ℓ ( 2 c − 1 α + 1 ) M e i θ , 1 + α 3 c ( c − 1 ) ( c − 2 ) ( N + 3 L ( c − 1 α + 1 ) + ( c ( c − 1 ) ( c − 2 ) α 3 + ℓ ( 3 c 2 − 6 c + 2 α 2 + 3 ( c − 1 ) α + 1 ) ) M e i θ ) ; z ) ∉ Ω ,$
where $z ∈ U$, $R e ( L e − i θ ) ≥ ℓ ( ℓ − 1 ) M$ and $R e ( N e − i θ ) ≥ 0$ for all real θ and $ℓ ∈ N \ { 1 }$.
Corollary 6.
Let $ϕ ∈ Ψ Γ [ Ω , M ]$. If $f ( z ) ∈ H$ satisfy the following conditions:
$1 z H α , β η , k ( f ( z ) ) − H α , β + 1 η , k ( f ( z ) ) ≤ | α c | ℓ M ,$
and
$ϕ H α , β + 1 η , k ( f ( z ) ) z , H α , β η , k ( f ( z ) ) z , H α , β − 1 η , k ( f ( z ) ) z , H α , β − 2 η , k ( f ( z ) ) z ; z ∈ Ω ,$
then
$H α , β + 1 η , k ( f ( z ) ) z − 1 < M .$
In the case $Ω = q ( U ) = ω : | ω − 1 | < M , ( M > 0 )$, we use notation $Φ Γ [ M ]$ to the class $Φ Γ [ Ω , M ]$.
Corollary 7.
Let $ϕ ∈ Ψ Γ [ Ω , M ]$. If $f ( z ) ∈ H$ satisfy the conditions (30) and
$ϕ H α , β η , k ( f ( z ) ) z , H α , β η + 1 , k ( f ( z ) ) z , H α , β η + 2 , k ( f ( z ) ) z , H α , β η + 3 , k ( f ( z ) ) z ; z − 1 < M ,$
then
$H α , β η , k ( f ( z ) ) z − 1 < M .$
Putting $ϕ ( a 1 , a 2 , a 3 , a 4 ; z ) = a 2 = 1 + ( k b ) b k + ℓ M e i θ$ in Corollary 7, we have the following corollary:
Corollary 8.
Let $M > 0$ and $b ∈ C \ Z 0 −$ with $R e ( b ) < − ℓ 2$, $ℓ ∈ N \ { 1 }$. If $f ( z ) ∈ H$ satisfy the condition (30), and also, if:
$H α , β η + 1 , k ( f ( z ) ) z − 1 < M ,$
then
$H α , β η , k ( f ( z ) ) z − 1 < M .$
Corollary 9.
Let $ϕ ∈ Ψ Γ [ Ω , M ]$. If $f ( z ) ∈ H$ satisfy the conditions (31) and
$ϕ H α , β + 1 η , k ( f ( z ) ) z , H α , β η , k ( f ( z ) ) z , H α , β − 1 η , k ( f ( z ) ) z , H α , β − 2 η , k ( f ( z ) ) z ; z − 1 < M ,$
then
$H α , β + 1 η , k ( f ( z ) ) z − 1 < M .$
Putting $ϕ ( a 1 , a 2 , a 3 , a 4 ; z ) = a 2 = 1 + ( α c ) c α + ℓ M e i θ$ in Corollary 9, we have the following corollary:
Corollary 10.
Let $M > 0$ and $c ∈ C \ Z 0 −$ with $R e ( c ) < − ℓ 2$, $ℓ ∈ N \ { 1 }$. If $f ( z ) ∈ H$ satisfy the conditions (31) and
$H α , β η , k ( f ( z ) ) z − 1 < M ,$
then
$H α , β + 1 η , k ( f ( z ) ) z − 1 < M .$
Corollary 11.
Let $ℓ ∈ N \ 1$, $M > 0$ and $b ∈ C \ Z 0 −$. If $f ( z ) ∈ H$ satisfies the condition (30) and
$1 z H α , β η + 1 , k ( f ( z ) ) − H α , β η , k ( f ( z ) ) ≤ | k b | ℓ M ,$
then
$H α , β η , k ( f ( z ) ) z − 1 < M .$
Proof.
Let $ϕ ( a 1 , a 2 , a 3 , a 4 ; z ) = a 2 − a 1 .$ Using Corollary 5 with $Ω = h ( U )$ and
$h ( z ) = | k b | ℓ M ( z ∈ U ) .$
Now we show that $ϕ ∈ Ψ Γ [ Ω , M ]$. Since the condition (30) is satisfied from the condition (32) and
$| ϕ ( 1 + M e i θ , 1 + ( k b ) b k + ℓ M e i θ , 1 + k 2 b ( b + 1 ) L + b ( b + 1 ) k 2 + ℓ ( 2 b + 1 k + 1 ) M e i θ , 1 + k 3 b ( b + 1 ) ( b + 2 ) ( N + 3 L ( b + 1 k + 1 ) + ( b ( b + 1 ) ( b + 2 ) k 3 + ℓ ( 3 b 2 + 6 b + 2 k 2 + 3 ( b + 1 ) k + 1 ) ) M e i θ ) ; z ) | = k ℓ M e i θ b = | k b | ℓ M ,$
then we have the Corollary 11. □
Corollary 12.
Let $ℓ ∈ N \ { 1 }$, $M > 0$ and $c ∈ C \ Z 0 −$. If $f ( z ) ∈ H$ satisfies the condition (31) and
$1 z H α , β η , k ( f ( z ) ) − H α , β + 1 η , k ( f ( z ) ) ≤ | α c | ℓ M ,$
then
$H α , β + 1 η , k ( f ( z ) ) z − 1 < M .$
Proof.
Let $ϕ ( a 1 , a 2 , a 3 , a 4 ; z ) = a 2 − a 1 .$ Using Corollary 6 with $Ω = h ( U )$ and
$h ( z ) = | α c | ℓ M ( z ∈ U ) .$
Now we show that $ϕ ∈ Ψ Γ [ Ω , M ]$. Since the condition (31) is satisfied from the condition (33) and
$| ϕ ( 1 + M e i θ , 1 + ( α c ) c α + ℓ M e i θ , 1 + α 2 c ( c − 1 ) L + c ( c − 1 ) α 2 + ℓ ( 2 c − 1 α + 1 ) M e i θ , 1 + α 3 c ( c − 1 ) ( c − 2 ) ( N + 3 L ( c − 1 α + 1 ) + ( c ( c − 1 ) ( c − 2 ) α 3 + ℓ ( 3 c 2 − 6 c + 2 α 2 + 3 ( c − 1 ) α + 1 ) ) M e i θ ) ; z ) | = α ℓ M e i θ c = | α c | ℓ M ,$
then the corollary is completed. □
Corollary 13.
Let $ℓ ∈ N \ 1$, $M > 0$ and $b ∈ C \ Z 0 −$. If $f ( z ) ∈ H$ satisfy the condition (30) and
$1 z H α , β η + 3 , k ( f ( z ) ) − H α , β η + 2 , k ( f ( z ) ) ≤ 2 k 3 b ( b + 1 ) ( b + 2 ) 2 b + 1 k + 3 + b ( b + 1 ) k 2 + 2 b + 1 k + 1 M ,$
then
$H α , β η , k ( f ( z ) ) z − 1 < M .$
Proof.
Let $ϕ ( a 1 , a 2 , a 3 , a 4 ; z ) = a 4 − a 3$. We use Corollary 5 with $Ω = h ( U )$
$h ( z ) = 2 k 3 b ( b + 1 ) ( b + 2 ) 2 b + 1 k + 3 + b ( b + 1 ) k 2 + 2 b + 1 k + 1 M z , ( z ∈ U ) .$
Now we show that $ϕ ∈ Ψ Γ [ Ω , M ]$. Since
$| ϕ ( 1 + M e i θ , 1 + ( k b ) b k + ℓ M e i θ , 1 + k 2 b ( b + 1 ) ( L + ( b ( b + 1 ) k 2 + ℓ ( 2 b + 1 k + 1 ) M e i θ , 1 + k 3 b ( b + 1 ) ( b + 2 ) ( N + 3 L ( b + 1 k + 1 ) + ( b ( b + 1 ) ( b + 2 ) k 3 + ℓ ( 3 b 2 + 6 b + 2 k 2 + 3 ( b + 1 ) k + 1 ) M e i θ ; z ) | = k 3 b ( b + 1 ) ( b + 2 ) N + ( 2 b + 1 k + 3 ) L + ℓ b ( b + 1 ) k 2 + 2 b + 1 k + 1 M e i θ = k 3 e i θ b ( b + 1 ) ( b + 2 ) N e − i θ + ( 2 b + 1 k + 3 ) L e − i θ + ℓ b ( b + 1 ) k 2 + 2 b + 1 k + 1 M ≥ k 3 b ( b + 1 ) ( b + 2 ) ( R e ( N e − i θ ) + 2 b + 1 k + 3 R e ( L e − i θ ) + ℓ b ( b + 1 ) k 2 + 2 b + 1 k + 1 M ) ≥ k 3 b ( b + 1 ) ( b + 2 ) ℓ ( ℓ − 1 ) M 2 b + 1 k + 3 + ℓ b ( b + 1 ) k 2 + 2 b + 1 k + 1 M ≥ 2 k 3 b ( b + 1 ) ( b + 2 ) 2 b + 1 k + 3 + b ( b + 1 ) k 2 + 2 b + 1 k + 1 M ,$
we complete the proof of Corollary 13. □
Corollary 14.
Let $ℓ ∈ N \ { 1 }$, $M > 0$ and $c ∈ C \ Z 0 −$. If $f ( z ) ∈ H$ satisfy the condition (31) and
$1 z H α , β − 2 η , k ( f ( z ) ) − H α , β − 1 η , k ( f ( z ) ) ≤ 2 α 3 c ( c − 1 ) ( c − 2 ) 2 c − 1 α + 3 + c ( c − 1 ) α 2 + 2 c − 1 α + 1 M ,$
then
$H α , β + 1 η , k ( f ( z ) ) z − 1 < M .$
Proof.
Let $ϕ ( a 1 , a 2 , a 3 , a 4 ; z ) = a 4 − a 3 .$ We use Corollary 6 with $Ω = h ( U )$
$h ( z ) = 2 α 3 c ( c − 1 ) ( c − 2 ) 2 c − 1 α + 3 + c ( c − 1 ) α 2 + 2 c − 1 α + 1 M z , ( z ∈ U ) .$
Now we show that $ϕ ∈ Ψ Γ [ Ω , M ]$. Since
$| ϕ ( 1 + M e i θ , 1 + ( α c ) c α + ℓ M e i θ , 1 + α 2 c ( c − 1 ) ( L + ( c ( c − 1 ) α 2 + ℓ ( 2 c − 1 α + 1 ) M e i θ , 1 + α 3 c ( c − 1 ) ( c − 2 ) ( N + 3 L ( c − 1 α + 1 ) + ( c ( c − 1 ) ( c − 2 ) α 3 + ℓ ( 3 c 2 − 6 c + 2 α 2 + 3 ( c − 1 ) α + 1 ) ) M e i θ ) ; z ) | = | α 3 c ( c − 1 ) ( c − 2 ) ( N + ( 2 c − 1 α + 3 ) L + ℓ c ( c − 1 ) α 2 + 2 c − 1 α + 1 M e i θ = α 3 e − i θ c ( c − 1 ) ( c − 2 ) N e − i θ + ( 2 c − 1 α + 3 ) L e − i θ + ℓ ( c ( c − 1 ) α 2 + 2 c − 1 α + 1 ) M ≥ α 3 c ( c − 1 ) ( c − 2 ) ( R e ( N e − i θ ) + 2 c − 1 α + 3 R e ( L e − i θ ) + ℓ c ( c − 1 ) α 2 + 2 c − 1 α + 1 M ) ≥ α 3 c ( c − 1 ) ( c − 2 ) ℓ ( ℓ − 1 ) M 2 c − 1 k + 3 + ℓ c ( c − 1 ) α 2 + 2 c − 1 α + 1 M ≥ 2 α 3 c ( c − 1 ) ( c − 2 ) 2 c − 1 α + 3 + c ( c − 1 ) α 2 + 2 c − 1 α + 1 M ,$
we complete the proof of Corollary 14. □

## 3. Third Order Differential Supordination with $H α , β η , k ( f ( z ) )$

Definition 11.
Let $Ω ⊆ C$ and $q ( z ) ∈ Q$. The class of admissible functions $Ψ Γ [ Ω , q ( z ) ]$ consists of those functions $ψ : C 4 × U ¯ ⟶ C$ that satisfy the admissibility condition
$ψ ( a 1 , a 2 , a 3 , a 4 ; z ) ∉ Ω ,$
whenever
$a 1 = q ( ζ ) , a 2 = k ζ q ′ ( ζ ) + b q ( ζ ) b m , R e ( b + 1 ) ( a 3 − a 1 ) k ( a 2 − a 1 ) − 2 b + 1 k ≤ 1 m R e ζ q ″ ( ζ ) q ′ ( ζ ) + 1 , R e ( ( b + 1 ) ( b + 2 ) ( a 4 − a 1 ) − 3 ( b + 1 ) ( b + k + 1 ) ( a 3 − a 1 ) k 2 ( a 2 − a 1 ) + 3 b ( b + 1 ) + 1 k 2 + 6 b + 3 k + 2 ) ≤ 1 m 2 R e ζ 2 q ‴ ( ζ ) q ′ ( ζ ) ,$
where $z ∈ U$; $ζ ∈ ∂ U$, $m ∈ N \ { 1 }$ and $b = η + k$.
Theorem 9.
Let $ϕ ∈ Ψ Γ ′ [ Ω , q ( z ) ]$. If $f ( z ) ∈ H$ and $H α , β η , k ( f ( z ) ) z ∈ Q 1$ satisfy the following conditions:
$R e z q ″ ( z ) q ′ ( z ) ≥ 0 , 1 z H α , β η + 1 , k ( f ( z ) ) − H α , β η , k ( f ( z ) ) q ′ ( z ) ≤ | k b | m , ϕ H α , β η , k ( f ( z ) ) z , H α , β η + 1 , k ( f ( z ) ) z , H α , β η + 2 , k ( f ( z ) ) z , H α , β η + 3 , k ( f ( z ) ) z ; z : z ∈ U ,$
are univalent, and
$Ω ⊂ { ϕ ( H α , β η , k ( f ( z ) ) z , H α , β η + 1 , k ( f ( z ) ) z , H α , β η + 2 , k ( f ( z ) ) z , H α , β η + 3 , k ( f ( z ) ) z ; z : z ∈ U ,$
then
$q ( z ) ≺ H α , β η , k ( f ( z ) ) z .$
Proof.
Let the functions $p ( z )$ and $ψ$ are defined by (9) and (13). Since $ϕ ∈ Ψ Γ ′ [ Ω , q ( z ) ]$, therefore (14) and (35) give
$Ω ⊂ ψ ( p ( z ) , z p ′ ( z ) , z 2 p ″ ( z ) , z 3 p ‴ ( z ) ; z ) .$
The admissible condition for $ϕ ∈ Ψ Γ ′ [ Ω , q ( z ) ]$ is equivalent to the admissible condition for $ψ$ in Definition 6 with $n = 2$. Therefore, $ψ ∈ Ψ Γ ′ [ Ω , q ( z ) ]$ and by using (34) and Theorem 2, we have $q ( z ) ≺ p ( z )$ which yields $q ( z ) ≺ H α , β η , k ( f ( z ) ) z .$ Therefore, we complete the proof of Theorem 9. □
Moreover, in a similar way, we can define the following:
Definition 12.
Let $Ω ⊆ C$ and $q ( z ) ∈ Q$. The class of admissible functions $Ψ Γ [ Ω , q ( z ) ]$ consists of those functions $ψ : C 4 × U ¯ ⟶ C$ that satisfy the admissibility condition:
$ψ ( a 1 , a 2 , a 3 , a 4 ; z ) ∉ Ω ,$
whenever
$a 1 = q ( ζ ) , a 2 = α ζ q ′ ( ζ ) + c q ( ζ ) c m , R e ( c − 1 ) ( a 3 − a 1 ) α ( a 2 − a 1 ) − 2 c − 1 α ≤ 1 m R e ζ q ″ ( ζ ) q ′ ( ζ ) + 1 , R e ( ( c − 1 ) ( c − 2 ) ( a 4 − a 1 ) − 3 ( c − 1 ) ( c + α − 1 ) ( a 3 − a 1 ) α 2 ( a 2 − a 1 ) + 3 c ( c − 1 ) + 1 α 2 + 6 c − 3 α + 2 ) ≤ 1 m 2 R e ζ 2 q ‴ ( ζ ) q ′ ( ζ ) ,$
where $z ∈ U$; $ζ ∈ ∂ U$, $m ∈ N \ { 1 }$ and $c = α + β$.
With the assist of Definition 12 and Theorem 4, we have the following theorem
Theorem 10.
Let $ϕ ∈ Ψ Γ ′ [ Ω , q ( z ) ]$. If $f ( z ) ∈ H$ and $H α , β + 1 η , k ( f ( z ) ) z ∈ Q 1$ satisfy the following conditions:
$R e z q ″ ( z ) q ′ ( z ) ≥ 0 , 1 z H α , β η , k ( f ( z ) ) − H α , β + 1 η , k ( f ( z ) ) q ′ ( z ) ≤ | α c | m , ϕ H α , β + 1 η , k ( f ( z ) ) z , H α , β η , k ( f ( z ) ) z , H α , β − 1 η , k ( f ( z ) ) z , H α , β − 2 η , k ( f ( z ) ) z ; z : z ∈ U ,$
are univalent, and
$Ω ⊂ ϕ H α , β + 1 η , k ( f ( z ) ) z , H α , β η , k ( f ( z ) ) z , H α , β − 1 η , k ( f ( z ) ) z , H α , β − 2 η , k ( f ( z ) ) z ; z : z ∈ U ,$
then
$q ( z ) ≺ H α , β + 1 η , k ( f ( z ) ) z .$
If $Ω ≠ C$ is a simply connected domain, then $Ω = h ( U )$ for some conformal mapping $h ( z )$ of $U$ onto $Ω .$ In this case, the class $Ψ Γ ′ [ h ( u ) , q ( z ) ]$ is written as $Ψ Γ ′ [ h , q ] .$ The following theorem is a direct consequence of Theorems 3 and 4.
Theorem 11.
Let $ϕ ∈ Ψ Γ ′ [ Ω , q ( z ) ]$ and $h ( z )$ be analytic function in $U$, and $f ( z ) ∈ H$ and $H α , β η , k ( f ( z ) ) z ∈ Q 1$ satisfy the condition (34). If
$ϕ H α , β η , k ( f ( z ) ) z , H α , β η + 1 , k ( f ( z ) ) z , H α , β η + 2 , k ( f ( z ) ) z , H α , β η + 3 , k ( f ( z ) ) z ; z : z ∈ U ,$
is univalent function in $U$, and
$h ( z ) ≺ ϕ H α , β η , k ( f ( z ) ) z , H α , β η + 1 , k ( f ( z ) ) z , H α , β η + 2 , k ( f ( z ) ) z , H α , β η + 3 , k ( f ( z ) ) z ; z ,$
then
$q ( z ) ≺ H α , β η , k ( f ( z ) ) z .$
Theorem 12.
Let $ϕ ∈ Ψ Γ ′ [ Ω , q ( z ) ]$ and $h ( z )$ be analytic function in $U$. If $f ( z ) ∈ H$ and $H α , β + 1 η , k ( f ( z ) ) z ∈ Q 1$ satisfy the condition (36). If
$ϕ H α , β + 1 η , k ( f ( z ) ) z , H α , β η , k ( f ( z ) ) z , H α , β − 1 η + 2 , k ( f ( z ) ) z , H α , β − 2 η , k ( f ( z ) ) z ; z : z ∈ U ,$
is univalent function in $U$ and
$h ( z ) ≺ ϕ H α , β + 1 η , k ( f ( z ) ) z , H α , β η , k ( f ( z ) ) z , H α , β − 1 η , k ( f ( z ) ) z , H α , β − 2 η , k ( f ( z ) ) z ; z ,$
then
$q ( z ) ≺ H α , β + 1 η , k ( f ( z ) ) z .$
Theorem 13.
Let $h ( z )$ be analytic function in $U$ and let $ψ : C 4 × U _ ⟶ C$ and ψ is given by (13). Suppose that the differential (26) has a solution $q ( z ) ∈ Q 1$, and $f ( z ) ∈ H$ satisfy the condition (34). If
$ϕ H α , β η , k ( f ( z ) ) z , H α , β η + 1 , k ( f ( z ) ) z , H α , β η + 2 , k ( f ( z ) ) z , H α , β η + 3 , k ( f ( z ) ) z ; z : z ∈ U ,$
is univalent function in $U$, and
$h ( z ) ≺ ϕ H α , β η , k ( f ( z ) ) z , H α , β η + 1 , k ( f ( z ) ) z , H α , β η + 2 , k ( f ( z ) ) z , H α , β η + 3 , k ( f ( z ) ) z ; z ,$
then
$q ( z ) ≺ H α , β η , k ( f ( z ) ) z .$
and $q ( z )$ is the best subordinant of relation (36).
Proof.
The proof is similar to Theorem 7 and it is being omitted here. □
Combining both Theorems 5 and 11, we have the following sandwich result:
Corollary 15.
Let $h 1 ( z )$ and $q 1 ( z )$ be analytic functions in $U$, also, let $h 2 ( z )$ be univalent in $U$, $q 2 ( z ) ∈ Q 1$ with $q 1 ( 0 ) = q 2 ( 0 ) = 1$ and $ϕ ∈ Ψ Γ [ Ω , q ( z ) ] ∩ Ψ Γ ′ [ Ω , q ( z ) ] .$ If $f ( z ) ∈ H$ and $H α , β η , k ( f ( z ) ) z ∈ Q 1 ∩ H$,
$ϕ H α , β η , k ( f ( z ) ) z , H α , β η + 1 , k ( f ( z ) ) z , H α , β η + 2 , k ( f ( z ) ) z , H α , β η + 3 , k ( f ( z ) ) z ; z : z ∈ U ,$
is univalent function in $U$, and the conditions (22) and (34) are satisfied, also let
$h 1 ( z ) ≺ ϕ H α , β η , k ( f ( z ) ) z , H α , β η + 1 , k ( f ( z ) ) z , H α , β η + 2 , k ( f ( z ) ) z , H α , β η + 3 , k ( f ( z ) ) z ; z ≺ h 2 ( z ) ,$
then
$q 1 ( z ) ≺ H α , β η , k ( f ( z ) ) z ≺ q 2 ( z ) .$
The proof of the following theorem is similar to Theorem 8; therefore, we omitted it.
Theorem 14.
Let $h ( z )$ be analytic function in $U$ and let $ψ : C 4 × U _ ⟶ C$ and ψ is given by (21). Suppose that the differential (28) has a solution $q ( z ) ∈ Q 1$.If $f ( z ) ∈ H$ satisfy the condition (36). If
$ϕ H α , β + 1 η , k ( f ( z ) ) z , H α , β η , k ( f ( z ) ) z , H α , β − 1 η , k ( f ( z ) ) z , H α , β − 2 η , k ( f ( z ) ) z ; z : z ∈ U ,$
is univalent function in $U$, and
$h ( z ) ≺ ϕ H α , β + 1 η , k ( f ( z ) ) z , H α , β η , k ( f ( z ) ) z , H α , β − 1 η , k ( f ( z ) ) z , H α , β − 2 η , k ( f ( z ) ) z ; z ,$
then
$q ( z ) ≺ H α , β + 1 η , k ( f ( z ) ) z .$
and $q ( z )$ is the best subordinant of (38).
By combining Theorems 8 and 12, we obtain the following sandwich type result.
Corollary 16.
Let $h 1 ( z )$ and $q 1 ( z )$ be analytic functions in $U$ and let $h 2 ( z )$ be univalent in $U$, $q 2 ( z ) ∈ Q 1$ with $q 1 ( 0 ) = q 2 ( 0 ) = 1$ and $ϕ ∈ Ψ Γ [ Ω , q ( z ) ] ∩ Ψ Γ ′ [ Ω , q ( z ) ]$. If $f ( z ) ∈ H$ and $H α , β + 1 η , k ( f ( z ) ) z ∈ Q 1 ∩ H$,
$ϕ H α , β + 1 η , k ( f ( z ) ) z , H α , β η , k ( f ( z ) ) z , H α , β − 1 η , k ( f ( z ) ) z , H α , β − 2 η , k ( f ( z ) ) z ; z : z ∈ U ,$
is univalent function in $U$, and the conditions (23) and (36) are satisfied; also let
$h 1 ( z ) ≺ ϕ H α , β + 1 η , k ( f ( z ) ) z , H α , β η , k ( f ( z ) ) z , H α , β − 1 η , k ( f ( z ) ) z , H α , β − 2 η , k ( f ( z ) ) z ; z ≺ h 2 ( z ) ,$
then
$q 1 ( z ) ≺ H α , β + 1 η , k ( f ( z ) ) z ≺ q 2 ( z ) .$

## 4. Conclusions

By using the method of third-order differential subordination and superordination, we obtained many interesting results concerning the subordination and superordination properties of analytic functions associated with the operator $H α , β η , k ( f )$.

## Author Contributions

Data curation, M.F.Y.; Formal analysis, M.F.Y.; Funding acquisition, M.F.Y.; Investigation, A.A.A.; Methodology, A.A.A.; Project administration, P.A.; Resources, M.F.Y.; Writing—original draft, A.A.A.; Writing—review and editing, P.A. All authors have read and agreed to the published version of the manuscript.

## Funding

This project was supported by the Deanship of Scientific Research at Prince Sattam Bin Abdulaziz University under the research project: 2019/01/10558.

## Conflicts of Interest

The authors declare no conflict of interest.

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Yassen, M.F.; Attiya, A.A.; Agarwal, P. Subordination and Superordination Properties for Certain Family of Analytic Functions Associated with Mittag–Leffler Function. Symmetry 2020, 12, 1724. https://doi.org/10.3390/sym12101724

AMA Style

Yassen MF, Attiya AA, Agarwal P. Subordination and Superordination Properties for Certain Family of Analytic Functions Associated with Mittag–Leffler Function. Symmetry. 2020; 12(10):1724. https://doi.org/10.3390/sym12101724

Chicago/Turabian Style

Yassen, Mansour F., Adel A. Attiya, and Praveen Agarwal. 2020. "Subordination and Superordination Properties for Certain Family of Analytic Functions Associated with Mittag–Leffler Function" Symmetry 12, no. 10: 1724. https://doi.org/10.3390/sym12101724

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