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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,
Adel A. Attiya
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
(This article belongs to the Special Issue Advanced Calculus in Problems with Symmetry)

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 = ζ q ( ζ ) , R e t s + 1 R e ζ q ( ζ ) q ( ζ ) + 1 ,
and
R e u s 2 R e ζ 2 q ( ζ ) q ( ζ ) ,
where z U ; ζ U \ E ( q ) and 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 ) , ( z U ; ζ U \ E ( q ) ; 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 = 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 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 ) 2 R e ζ 2 q ( ζ ) q ( ζ ) ,
where z U ; ζ U \ E ( q ) , 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 | ,
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 ) | ,
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 = 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 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 ) 2 R e ζ 2 q ( ζ ) q ( ζ ) ,
where z U ; ζ U \ E ( q ) , 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 | ,
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 ) ,
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 | ,
( 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 | ,
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 | ,
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 | ,
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 | ,
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 | ,
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 + M e i θ , 1 + k 2 b ( b + 1 ) L + b ( b + 1 ) k 2 + ( 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 + ( 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 ) ( 1 ) M and R e ( N e i θ ) 0 for all real θ and 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 | 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 + M e i θ , 1 + α 2 c ( c 1 ) L + c ( c 1 ) α 2 + ( 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 + ( 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 ) ( 1 ) M and R e ( N e i θ ) 0 for all real θ and 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 | 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 + 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 ) < 2 , 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 + 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 ) < 2 , 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 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 | 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 | 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 + M e i θ , 1 + k 2 b ( b + 1 ) L + b ( b + 1 ) k 2 + ( 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 + ( 3 b 2 + 6 b + 2 k 2 + 3 ( b + 1 ) k + 1 ) ) M e i θ ) ; z ) | = k M e i θ b = | k b | M ,
then we have the Corollary 11. □
Corollary 12.
Let 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 | 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 | 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 + M e i θ , 1 + α 2 c ( c 1 ) L + c ( c 1 ) α 2 + ( 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 + ( 3 c 2 6 c + 2 α 2 + 3 ( c 1 ) α + 1 ) ) M e i θ ) ; z ) | = α M e i θ c = | α c | M ,
then the corollary is completed. □
Corollary 13.
Let 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 + M e i θ , 1 + k 2 b ( b + 1 ) ( L + ( b ( b + 1 ) k 2 + ( 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 + ( 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 + 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 + 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 ) + b ( b + 1 ) k 2 + 2 b + 1 k + 1 M ) k 3 b ( b + 1 ) ( b + 2 ) ( 1 ) M 2 b + 1 k + 3 + 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 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 + M e i θ , 1 + α 2 c ( c 1 ) ( L + ( c ( c 1 ) α 2 + ( 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 + ( 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 + 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 + ( 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 ) + c ( c 1 ) α 2 + 2 c 1 α + 1 M ) α 3 c ( c 1 ) ( c 2 ) ( 1 ) M 2 c 1 k + 3 + 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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